www.vnmath.com

1

BI DNG HC SINH GII TON 9 www.vnmath.com

www.VNMATH.com

www.vnmath.com

2

CHUYN 1 : A THC

B. CC PHNG PHP V BI TP: I. TCH MT HNG T THNH NHIU HNG T: * nh l b sung: + a thc f(x) c nghim hu t th c dng p/q trong p l c ca h s t do, q l c dng ca h s cao nht + Nu f(x) c tng cc h s bng 0 th f(x) c mt nhn t l x 1 + Nu f(x) c tng cc h s ca cc hng t bc chn bng tng cc h s ca cc hng t bc l th f(x) c mt nhn t l x + 1 + Nu a l nghim nguyn ca f(x) v f(1); f(- 1) khc 0 th f(1)

a - 1 v f(-1)

a + 1 u l s

nguyn. nhanh chng loi tr nghim l c ca h s t do 1. V d 1: 3x2 8x + 4 Cch 1: Tch hng t th 2 3x2 8x + 4 = 3x2 6x 2x + 4 = 3x(x 2) 2(x 2) = (x 2)(3x 2) Cch 2: Tch hng t th nht: 3x2 8x + 4 = (4x2 8x + 4) - x2 = (2x 2)2 x2 = (2x 2 + x)(2x 2 x) = (x 2)(3x 2) 2. V d 2: x3 x2 - 4 Ta nhn thy nghim ca f(x) nu c th x = 1; 2; 4 , ch c f(2) = 0 nn x = 2 l nghim ca f(x) nn f(x) c mt nhn t l x 2. Do ta tch f(x) thnh cc nhm c xut hin mt nhn t l x 2 Cch 1: x3 - x2 4 = 3 2 2 2x 2x x 2x 2x 4 x x 2 x(x 2) 2(x 2) = 2x 2 x x 2 Cch 2: 3 2 3 2 3 2x x 4 x 8 x 4 x 8 x 4

2(x 2)(x 2x 4) (x 2)(x 2) = 2 2x 2 x 2x 4 (x 2) (x 2)(x x 2) 3. V d 3: f(x) = 3x3 7x2 + 17x 5 Nhn xt: 1, 5 khng l nghim ca f(x), nh vy f(x) khng c nghim nguyn. Nn f(x) nu c nghim th l nghim hu t Ta nhn thy x = 1

3 l nghim ca f(x) do f(x) c mt nhn t l 3x 1. Nn

f(x) = 3x3 7x2 + 17x 5 = 3 2 2 3 2 23x x 6x 2x 15x 5 3x x 6x 2x 15x 5 = 2 2x (3x 1) 2x(3x 1) 5(3x 1) (3x 1)(x 2x 5) V 2 2 2x 2x 5 (x 2x 1) 4 (x 1) 4 0 vi mi x nn khng phn tch c thnh nhn t na 4. V d 4: x3 + 5x2 + 8x + 4 Nhn xt: Tng cc h s ca cc hng t bc chn bng tng cc h s ca cc hng t bc l nn a thc c mt nhn t l x + 1 x3 + 5x2 + 8x + 4 = (x3 + x2 ) + (4x2 + 4x) + (4x + 4) = x2(x + 1) + 4x(x + 1) + 4(x + 1)

www.VNMATH.com

www.vnmath.com

3

= (x + 1)(x2 + 4x + 4) = (x + 1)(x + 2)2 5. V d 5: f(x) = x5 2x4 + 3x3 4x2 + 2 Tng cc h s bng 0 th nn a thc c mt nhn t l x 1, chia f(x) cho (x 1) ta c: x5 2x4 + 3x3 4x2 + 2 = (x 1)(x4 - x3 + 2 x2 - 2 x - 2) V x4 - x3 + 2 x2 - 2 x - 2 khng c nghim nguyn cng khng c nghim hu t nn khng phn tch c na 6.V d 6: x4 + 1997x2 + 1996x + 1997 = (x4 + x2 + 1) + (1996x2 + 1996x + 1996) = (x2 + x + 1)(x2 - x + 1) + 1996(x2 + x + 1)= (x2 + x + 1)(x2 - x + 1 + 1996) = (x2 + x + 1)(x2 - x + 1997) 7. V d 7: x2 - x - 2001.2002 = x2 - x - 2001.(2001 + 1) = x2 - x 20012 - 2001 = (x2 20012) (x + 2001) = (x + 2001)(x 2002) II. THM , BT CNG MT HNG T: 1. Thm, bt cng mt s hng t xut hin hiu hai bnh phng: a) V d 1: 4x4 + 81 = 4x4 + 36x2 + 81 - 36x2 = (2x2 + 9)2 36x2 = (2x2 + 9)2 (6x)2 = (2x2 + 9 + 6x)(2x2 + 9 6x) = (2x2 + 6x + 9 )(2x2 6x + 9) b) V d 2: x8 + 98x4 + 1 = (x8 + 2x4 + 1 ) + 96x4 = (x4 + 1)2 + 16x2(x4 + 1) + 64x4 - 16x2(x4 + 1) + 32x4 = (x4 + 1 + 8x2)2 16x2(x4 + 1 2x2) = (x4 + 8x2 + 1)2 - 16x2(x2 1)2 = (x4 + 8x2 + 1)2 - (4x3 4x )2 = (x4 + 4x3 + 8x2 4x + 1)(x4 - 4x3 + 8x2 + 4x + 1) 2. Thm, bt cng mt s hng t xut hin nhn t chung a) V d 1: x7 + x2 + 1 = (x7 x) + (x2 + x + 1 ) = x(x6 1) + (x2 + x + 1 ) = x(x3 - 1)(x3 + 1) + (x2 + x + 1 ) = x(x 1)(x2 + x + 1 ) (x3 + 1) + (x2 + x + 1) = (x2 + x + 1)[x(x 1)(x3 + 1) + 1] = (x2 + x + 1)(x5 x4 + x2 - x + 1) b) V d 2: x7 + x5 + 1 = (x7 x ) + (x5 x2 ) + (x2 + x + 1) = x(x3 1)(x3 + 1) + x2(x3 1) + (x2 + x + 1) = (x2 + x + 1)(x 1)(x4 + x) + x2 (x 1)(x2 + x + 1) + (x2 + x + 1) = (x2 + x + 1)[(x5 x4 + x2 x) + (x3 x2 ) + 1] = (x2 + x + 1)(x5 x4 + x3 x + 1) * Ghi nh: Cc a thc c dng x3m + 1 + x3n + 2 + 1 nh: x7 + x2 + 1 ; x7 + x5 + 1 ; x8 + x4 + 1 ; x5 + x + 1 ; x8 + x + 1 ; u c nhn t chung l x2 + x + 1 III. T BIN PH: 1. V d 1: x(x + 4)(x + 6)(x + 10) + 128 = [x(x + 10)][(x + 4)(x + 6)] + 128 = (x2 + 10x) + (x2 + 10x + 24) + 128 t x2 + 10x + 12 = y, a thc c dng (y 12)(y + 12) + 128 = y2 144 + 128 = y2 16 = (y + 4)(y 4) = ( x2 + 10x + 8 )(x2 + 10x + 16 ) = (x + 2)(x + 8)( x2 + 10x + 8 ) 2. V d 2: A = x4 + 6x3 + 7x2 6x + 1 Gi s x 0 ta vit x4 + 6x3 + 7x2 6x + 1 = x2 ( x2 + 6x + 7 2

6 1 + x x

) = x2 [(x2 + 21 x

) + 6(x - 1 x

) + 7 ]

t x - 1 x

= y th x2 + 21 x

= y2 + 2, do

www.VNMATH.com

www.vnmath.com

4

A = x2(y2 + 2 + 6y + 7) = x2(y + 3)2 = (xy + 3x)2 = [x(x - 1 x

)2 + 3x]2 = (x2 + 3x 1)2

* Ch : V d trn c th gii bng cch p dng hng ng thc nh sau: A = x4 + 6x3 + 7x2 6x + 1 = x4 + (6x3 2x2 ) + (9x2 6x + 1 ) = x4 + 2x2(3x 1) + (3x 1)2 = (x2 + 3x 1)2 3. V d 3: A = 2 2 2 2 2(x y z )(x y z) (xy yz+zx) = 2 2 2 2 2 2 2(x y z ) 2(xy yz+zx) (x y z ) (xy yz+zx) t 2 2 2x y z = a, xy + yz + zx = b ta c A = a(a + 2b) + b2 = a2 + 2ab + b2 = (a + b)2 = ( 2 2 2x y z + xy + yz + zx)2 4. V d 4: B = 4 4 4 2 2 2 2 2 2 2 2 42( ) ( ) 2( )( ) ( )x y z x y z x y z x y z x y z t x4 + y4 + z4 = a, x2 + y2 + z2 = b, x + y + z = c ta c: B = 2a b2 2bc2 + c4 = 2a 2b2 + b2 - 2bc2 + c4 = 2(a b2) + (b c2)2 Ta li c: a b2 = - 2( 2 2 2 2 2 2x y y z z x ) v b c2 = - 2(xy + yz + zx) Do : B = - 4( 2 2 2 2 2 2x y y z z x ) + 4 (xy + yz + zx)2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 24x y 4y z 4z x 4x y 4y z 4z x 8x yz 8xy z 8xyz8xyz(x y z)

5. V d 5: 3 3 3 3(a b c) 4(a b c ) 12abc t a + b = m, a b = n th 4ab = m2 n2 a3 + b3 = (a + b)[(a b)2 + ab] = m(n2 +

2 2m - n4

). Ta c:

C = (m + c)3 4. 3 2

3 2 2m + 3mn 4c 3c(m - n )4

= 3( - c3 +mc2 mn2 + cn2) = 3[c2(m - c) - n2(m - c)] = 3(m - c)(c - n)(c + n) = 3(a + b - c)(c + a - b)(c - a + b) IV. PHNG PHP H S BT NH: 1. V d 1: x4 - 6x3 + 12x2 - 14x + 3 Nhn xt: cc s 1, 3 khng l nghim ca a thc, a thc khng c nghim nguyn cng khng c nghim hu t Nh vy nu a thc phn tch c thnh nhn t th phi c dng (x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd

ng nht a thc ny vi a thc cho ta c: a c 6ac b d 12ad bc 14bd 3

Xt bd = 3 vi b, d Z, b 1, 3 vi b = 3 th d = 1 h iu kin trn tr thnh a c 6ac 8 2c 8 c 4a 3c 14 ac 8 a 2bd 3

Vy: x4 - 6x3 + 12x2 - 14x + 3 = (x2 - 2x + 3)(x2 - 4x + 1)

www.VNMATH.com

www.vnmath.com

5

2. V d 2: 2x4 - 3x3 - 7x2 + 6x + 8 Nhn xt: a thc c 1 nghim l x = 2 nn c tha s l x - 2 do ta c: 2x4 - 3x3 - 7x2 + 6x + 8 = (x - 2)(2x3 + ax2 + bx + c)

= 2x4 + (a - 4)x3 + (b - 2a)x2 + (c - 2b)x - 2c a 4 3

a 1b 2a 7

b 5c 2b 6

c 42c 8

Suy ra: 2x4 - 3x3 - 7x2 + 6x + 8 = (x - 2)(2x3 + x2 - 5x - 4) Ta li c 2x3 + x2 - 5x - 4 l a thc c tng h s ca cc hng t bc l v bc chn bng nhau nn c 1 nhn t l x + 1 nn 2x3 + x2 - 5x - 4 = (x + 1)(2x2 - x - 4) Vy: 2x4 - 3x3 - 7x2 + 6x + 8 = (x - 2)(x + 1)(2x2 - x - 4) 3. V d 3: 12x2 + 5x - 12y2 + 12y - 10xy - 3 = (a x + by + 3)(cx + dy - 1)

= acx2 + (3c - a)x + bdy2 + (3d - b)y + (bc + ad)xy 3

ac 12a 4

bc ad 10c 3

3c a 5b 6

bd 12d 2

3d b 12

12x2 + 5x - 12y2 + 12y - 10xy - 3 = (4 x - 6y + 3)(3x + 2y - 1) BI TP: Phn tch cc a thc sau thnh nhn t:

www.vnmath.com

1) x3 - 7x + 6 2) x3 - 9x2 + 6x + 16 3) x3 - 6x2 - x + 30 4) 2x3 - x2 + 5x + 3 5) 27x3 - 27x2 + 18x - 4 6) x2 + 2xy + y2 - x - y - 12 7) (x + 2)(x +3)(x + 4)(x + 5) - 24 8) 4x4 - 32x2 + 1 9) 3(x4 + x2 + 1) - (x2 + x + 1)2

10) 64x4 + y4 11) a6 + a4 + a2b2 + b4 - b6 12) x3 + 3xy + y3 - 1 13) 4x4 + 4x3 + 5x2 + 2x + 1 14) x8 + x + 1 15) x8 + 3x4 + 4 16) 3x2 + 22xy + 11x + 37y + 7y2 +10 17) x4 - 8x + 63

www.VNMATH.com

www.vnmath.com

6

CHUYN 2 - LU THA BC N CA MT NH THC

B. KIN THC V BI TP VN DNG: I. Mt s hng ng thc tng qut: 1. an - bn = (a - b)(an - 1 + an - 2 b + an - 3 b2 + + abn - 2 + bn - 1 ) 2. an + bn = (a + b) ( an - 1 - an - 2b + an - 3b2 - - abn - 2 + bn - 1 ) 3. Nh thc Niutn: (a + b)n = an + 1nC an - 1 b + 2nC an - 2 b2 + + n 1nC ab n - 1 + bn Trong : k n n(n - 1)(n - 2)...[n - (k - 1)]C 1.2.3...k : T hp chp k ca n phn t II. Cch xc nh h s ca khai trin Niutn: 1. Cch 1: Dng cng thc k n n(n - 1)(n - 2)...[n - (k - 1)]C k ! Chng hn h s ca hng t a4b3 trong khai trin ca (a + b)7 l 47 7.6.5.4 7.6.5.4C 354! 4.3.2.1 Ch : a) k n

n !C n!(n - k) !

vi quy c 0! = 1 47 7! 7.6.5.4.3.2.1C 354!.3! 4.3.2.1.3.2.1 b) Ta c: k nC = k - 1 nC nn 4 37 7

7.6.5.C C 353!

2. Cch 2: Dng tam gic Patxcan

nh 1 Dng 1(n = 1) 1 1 Dng 2(n = 1) 1 2 1 Dng 3(n = 3) 1 3 3 1 Dng 4(n = 4) 1 4 6 4 1 Dng 5(n = 5) 1 5 10 10 5 1 Dng 6(n = 6) 1 6 15 20 15 6 1

Trong tam gic ny, hai cnh bn gm cc s 1; dng k + 1 c thnh lp t dng k (k 1), chng hn dng 2 (n = 2) ta c 2 = 1 + 1, dng 3 (n = 3): 3 = 2 + 1, 3 = 1 + 2 dng 4 (n = 4): 4 = 1 + 3, 6 = 3 + 3, 4 = 3 + 1, Vi n = 4 th: (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 Vi n = 5 th: (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Vi n = 6 th: (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2 b4 + 6ab5 + b6 3. Cch 3: Tm h s ca hng t ng sau theo cc h s ca hng t ng trc: a) H s ca hng t th nht bng 1 b) Mun c h s ca ca hng t th k + 1, ta ly h s ca hng t th k nhn vi s m ca bin trong hng t th k ri chia cho k Chng hn: (a + b)4 = a4 + 1.41 a

3b + 4.32

a2b2 + 4.3.22.3

ab3 + 4.3.2.2.3.4

b5

Ch rng: cc h s ca khai trin Niutn c tnh i xng qua hng t ng gia, ngha l cc hng t cch u hai hng t u v cui c h s bng nhau (a + b)n = an + nan -1b + n(n - 1)

1.2an - 2b2 + + n(n - 1)

1.2a2bn - 2 + nan - 1bn - 1 + bn

www.VNMATH.com

www.vnmath.com

7

III. V d: 1. V d 1: phn tch a thc sau thnh nhn t a) A = (x + y)5 - x5 - y5 Cch 1: khai trin (x + y)5 ri rt gn A A = (x + y)5 - x5 - y5 = ( x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5) - x5 - y5 = 5x4y + 10x3y2 + 10x2y3 + 5xy4 = 5xy(x3 + 2x2y + 2xy2 + y3) = 5xy [(x + y)(x2 - xy + y2) + 2xy(x + y)] = 5xy(x + y)(x2 + xy + y2) Cch 2: A = (x + y)5 - (x5 + y5) x5 + y5 chia ht cho x + y nn chia x5 + y5 cho x + y ta c: x5 + y5 = (x + y)(x4 - x3y + x2y2 - xy3 + y4) nn A c nhn t chung l (x + y), t (x + y) lm nhn t chung, ta tm c nhn t cn li b) B = (x + y)7 - x7 - y7 = (x7+7x6y +21x5y2 + 35x4y3 +35x3y4 +21x2y5 7xy6 + y7) - x7 - y7 = 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 = 7xy[(x5 + y5 ) + 3(x4y + xy4) + 5(x3y2 + x2y3 )] = 7xy {[(x + y)(x4 - x3y + x2y2 - xy3 + y4) ] + 3xy(x + y)(x2 - xy + y2) + 5x2y2(x + y)} = 7xy(x + y)[x4 - x3y + x2y2 - xy3 + y4 + 3xy(x2 + xy + y2) + 5x2y2 ] = 7xy(x + y)[x4 - x3y + x2y2 - xy3 + y4 + 3x3y - 3x2y2 + 3xy3 + 5x2y2 ] = 7xy(x + y)[(x4 + 2x2y2 + y4) + 2xy (x2 + y2) + x2y2 ] = 7xy(x + y)(x2 + xy + y2 )2 V d 2:Tm tng h s cc a thc c c sau khi khai trin a) (4x - 3)4 Cch 1: Theo cnh thc Niu tn ta c: (4x - 3)4 = 4.(4x)3.3 + 6.(4x)2.32 - 4. 4x. 33 + 34 = 256x4 - 768x3 + 864x2 - 432x + 81 Tng cc h s: 256 - 768 + 864 - 432 + 81 = 1 b) Cch 2: Xt ng thc (4x - 3)4 = c0x4 + c1x3 + c2x2 + c3x + c4 Tng cc h s: c0 + c1 + c2 + c3 + c4 Thay x = 1 vo ng thc trn ta c: (4.1 - 3)4 = c0 + c1 + c2 + c3 + c4 Vy: c0 + c1 + c2 + c3 + c4 = 1 * Ghi ch: Tng cc h s khai trin ca mt nh thc, mt a thc bng gi tr ca a thc ti x = 1 C. BI TP: Bi 1: Phn tch thnh nhn t a) (a + b)3 - a3 - b3 b) (x + y)4 + x4 + y4 Bi 2: Tm tng cc h s c c sau khi khai trin a thc a) (5x - 2)5 b) (x2 + x - 2)2010 + (x2 - x + 1)2011

www.vnmath.com

www.VNMATH.com

www.vnmath.com

8

CHUYN 3 - CC BI TON V S CHIA HT CA S

NGUYN B.KIN THC V CC BI TON: I. Dng 1: Chng minh quan h chia ht 1. Kin thc: * chng minh A(n) chia ht cho mt s m ta phn tch A(n) thnh nhn t c mt nhn t lm hoc bi ca m, nu m l hp s th ta li phn tch n thnh nhn t c cc oi mt nguyn t cng nhau, ri chng minh A(n) chia ht cho cc s * Ch : + Vi k s nguyn lin tip bao gi cng tn ti mt bi ca k + Khi chng minh A(n) chia ht cho m ta xt mi trng hp v s d khi chia A(n) cho m + Vi mi s nguyn a, b v s t nhin n th: 2. Bi tp: 2. Cc bi ton Bi 1: chng minh rng a) 251 - 1 chia ht cho 7 b) 270 + 370 chia ht cho 13 c) 1719 + 1917 chi ht cho 18 d) 3663 - 1 chia ht cho 7 nhng khng chia ht cho 37 e) 24n -1 chia ht cho 15 vi n N Gii a) 251 - 1 = (23)17 - 1 23 - 1 = 7 b) 270 + 370 (22)35 + (32)35 = 435 + 935 4 + 9 = 13 c) 1719 + 1917 = (1719 + 1) + (1917 - 1) 1719 + 1 17 + 1 = 18 v 1917 - 1 19 - 1 = 18 nn (1719 + 1) + (1917 - 1) hay 1719 + 1917 18 d) 3663 - 1 36 - 1 = 35 7 3663 - 1 = (3663 + 1) - 2 chi cho 37 d - 2 e) 2 4n - 1 = (24) n - 1 24 - 1 = 15 Bi 2: chng minh rng a) n5 - n chia ht cho 30 vi n N ; b) n4 -10n2 + 9 chia ht cho 384 vi mi n l n Z c) 10n +18n -28 chia ht cho 27 vi n N ; Gii: a) n5 - n = n(n4 - 1) = n(n - 1)(n + 1)(n2 + 1) = (n - 1).n.(n + 1)(n2 + 1) chia ht cho 6 v (n - 1).n.(n+1) l tch ca ba s t nhin lin tip nn chia ht cho 2 v 3 (*) Mt khc n5 - n = n(n2 - 1)(n2 + 1) = n(n2 - 1).(n2 - 4 + 5) = n(n2 - 1).(n2 - 4 ) + 5n(n2 - 1) = (n - 2)(n - 1)n(n + 1)(n + 2) + 5n(n2 - 1) V (n - 2)(n - 1)n(n + 1)(n + 2) l tch ca 5 s t nhin lin tip nn chia ht cho 5

+) an - bn chia ht cho a - b (a - b) +) a2n + 1 + b2n + 1 chia ht cho a + b + (a + b)n = B(a) + bn

+) (a + 1)n l BS(a )+ 1

+)(a - 1)2n l B(a) + 1

+) (a - 1)2n + 1 l B(a) - 1

www.VNMATH.com

www.vnmath.com

9

5n(n2 - 1) chia ht cho 5 Suy ra (n - 2)(n - 1)n(n + 1)(n + 2) + 5n(n2 - 1) chia ht cho 5 (**) T (*) v (**) suy ra pcm b) t A = n4 -10n2 + 9 = (n4 -n2 ) - (9n2 - 9) = (n2 - 1)(n2 - 9) = (n - 3)(n - 1)(n + 1)(n + 3) V n l nn t n = 2k + 1 (k Z) th A = (2k - 2).2k.(2k + 2)(2k + 4) = 16(k - 1).k.(k + 1).(k + 2) A chia ht cho 16 (1) V (k - 1).k.(k + 1).(k + 2) l tch ca 4 s nguyn lin tip nn A c cha bi ca 2, 3, 4 nn A l bi ca 24 hay A chia ht cho 24 (2) T (1) v (2) suy ra A chia ht cho 16. 24 = 384 c) 10 n +18n -28 = ( 10 n - 9n - 1) + (27n - 27) + Ta c: 27n - 27 27 (1) + 10 n - 9n - 1 = [(

n

9...9 + 1) - 9n - 1] = n

9...9 - 9n = 9( n

1...1 - n) 27 (2)

v 9 9 v n

1...1 - n 3 do n

1...1 - n l mt s c tng cc ch s chia ht cho 3 T (1) v (2) suy ra pcm 3. Bi 3: Chng minh rng vi mi s nguyn a th a) a3 - a chia ht cho 3 b) a7 - a chia ht cho 7 Gii a) a3 - a = a(a2 - 1) = (a - 1) a (a + 1) l tch ca ba s nguyn lin tip nn tn ti mt s l bi ca 3 nn (a - 1) a (a + 1) chia ht cho 3 b) ) a7 - a = a(a6 - 1) = a(a2 - 1)(a2 + a + 1)(a2 - a + 1) Nu a = 7k (k Z) th a chia ht cho 7 Nu a = 7k + 1 (k Z) th a2 - 1 = 49k2 + 14k chia ht cho 7 Nu a = 7k + 2 (k Z) th a2 + a + 1 = 49k2 + 35k + 7 chia ht cho 7 Nu a = 7k + 3 (k Z) th a2 - a + 1 = 49k2 + 35k + 7 chia ht cho 7 Trong trng hp no cng c mt tha s chia ht cho 7 Vy: a7 - a chia ht cho 7 Bi 4: Chng minh rng A = 13 + 23 + 33 + ...+ 1003 chia ht cho B = 1 + 2 + 3 + ... + 100 Gii Ta c: B = (1 + 100) + (2 + 99) + ...+ (50 + 51) = 101. 50 chng minh A chia ht cho B ta chng minh A chia ht cho 50 v 101 Ta c: A = (13 + 1003) + (23 + 993) + ... +(503 + 513) = (1 + 100)(12 + 100 + 1002) + (2 + 99)(22 + 2. 99 + 992) + ... + (50 + 51)(502 + 50. 51 + 512) = 101(12 + 100 + 1002 + 22 + 2. 99 + 992 + ... + 502 + 50. 51 + 512) chia ht cho 101 (1) Li c: A = (13 + 993) + (23 + 983) + ... + (503 + 1003) Mi s hng trong ngoc u chia ht cho 50 nn A chia ht cho 50 (2) T (1) v (2) suy ra A chia ht cho 101 v 50 nn A chi ht cho B Bi tp v nh Chng minh rng: a) a5 a chia ht cho 5 b) n3 + 6n2 + 8n chia ht cho 48 vi mi n chn c) Cho a l s nguyn t ln hn 3. Cmr a2 1 chia ht cho 24

www.VNMATH.com

www.vnmath.com

10

d) Nu a + b + c chia ht cho 6 th a3 + b3 + c3 chia ht cho 6 e) 20092010 khng chia ht cho 2010 f) n2 + 7n + 22 khng chia ht cho 9 Dng 2: Tm s d ca mt php chia Bi 1: Tm s d khi chia 2100 a)cho 9, b) cho 25, c) cho 125 Gii a) Lu tha ca 2 st vi bi ca 9 l 23 = 8 = 9 - 1 Ta c : 2100 = 2. (23)33 = 2.(9 - 1)33 = 2.[B(9) - 1] = B(9) - 2 = B(9) + 7 Vy: 2100 chia cho 9 th d 7 b) Tng t ta c: 2100 = (210)10 = 102410 = [B(25) - 1]10 = B(25) + 1 Vy: 2100 chia chop 25 th d 1 c)S dng cng thc Niutn: 2100 = (5 - 1)50 = (550 - 5. 549 + +

50.492

. 52 - 50 . 5 ) + 1

Khng k phn h s ca khai trin Niutn th 48 s hng u cha tha s 5 vi s m ln hn hoc bng 3 nn u chia ht cho 53 = 125, hai s hng tip theo: 50.49

2. 52 - 50.5

cng chia ht cho 125 , s hng cui cng l 1 Vy: 2100 = B(125) + 1 nn chia cho 125 th d 1 Bi 2: Vit s 19951995 thnh tng ca cc s t nhin . Tng cc lp phng chia cho 6 th d bao nhiu? Gii t 19951995 = a = a1 + a2 + + an. Gi 3 3 3 31 2 3 nS a a + a + ...+ a = 3 3 3 31 2 3 na a + a + ...+ a + a - a = (a1 3 - a1) + (a2 3 - a2) + + (an 3 - an) + a Mi du ngoc u chia ht cho 6 v mi du ngoc l tch ca ba s t nhin lin tip. Ch cn tm s d khi chia a cho 6 1995 l s l chia ht cho 3, nn a cng l s l chia ht cho 3, do chia cho 6 d 3 Bi 3: Tm ba ch s tn cng ca 2100 vit trong h thp phn gii Tm 3 ch s tn cng l tm s d ca php chia 2100 cho 1000 Trc ht ta tm s d ca php chia 2100 cho 125 Vn dng bi 1 ta c 2100 = B(125) + 1 m 2100 l s chn nn 3 ch s tn cng ca n ch c th l 126, 376, 626 hoc 876 Hin nhin 2100 chia ht cho 8 v 2100 = 1625 chi ht cho 8 nn ba ch s tn cng ca n chia ht cho 8 trong cc s 126, 376, 626 hoc 876 ch c 376 chia ht cho 8 Vy: 2100 vit trong h thp phn c ba ch s tn cng l 376 Tng qut: Nu n l s chn khng chia ht cho 5 th 3 ch s tn cng ca n l 376 Bi 4: Tm s d trong php chia cc s sau cho 7 a) 2222 + 5555 b)31993

www.VNMATH.com

www.vnmath.com

11

c) 19921993 + 19941995 d) 193023 Gii a) ta c: 2222 + 5555 = (21 + 1)22 + (56 1)55 = (BS 7 +1)22 + (BS 7 1)55 = BS 7 + 1 + BS 7 - 1 = BS 7 nn 2222 + 5555 chia 7 d 0 b) Lu tha ca 3 st vi bi ca 7 l 33 = BS 7 1 Ta thy 1993 = BS 6 + 1 = 6k + 1, do : 31993 = 3 6k + 1 = 3.(33)2k = 3(BS 7 1)2k = 3(BS 7 + 1) = BS 7 + 3 c) Ta thy 1995 chia ht cho 7, do : 19921993 + 19941995 = (BS 7 3)1993 + (BS 7 1)1995 = BS 7 31993 + BS 7 1 Theo cu b ta c 31993 = BS 7 + 3 nn 19921993 + 19941995 = BS 7 (BS 7 + 3) 1 = BS 7 4 nn chia cho 7 th d 3 d) 193023 = 32860 = 33k + 1 = 3.33k = 3(BS 7 1) = BS 7 3 nn chia cho 7 th d 4 Bi tp v nh Tm s d khi: a) 21994 cho 7 b) 31998 + 51998 cho 13 c) A = 13 + 23 + 33 + ...+ 993 chia cho B = 1 + 2 + 3 + ... + 99 Dng 3: Tm iu kin xy ra quan h chia ht Bi 1: Tm n Z gi tr ca biu thc A = n3 + 2n2 - 3n + 2 chia ht cho gi tr ca biu thc B = n2 - n Gii Chia A cho B ta c: n3 + 2n2 - 3n + 2 = (n + 3)(n2 - n) + 2 A chia ht cho B th 2 phi chia ht cho n2 - n = n(n - 1) do 2 chia ht cho n, ta c:

n 1 - 1 2 - 2 n - 1 0 - 2 1 - 3

n(n - 1) 0 2 2 6 loi loi

Vy: gi tr ca biu thc A = n3 + 2n2 - 3n + 2 chia ht cho gi tr ca biu thc B = n2 - n th n 1;2 Bi 2: a) Tm n N n5 + 1 chia ht cho n3 + 1 b) Gii bi ton trn nu n Z Gii Ta c: n5 + 1 n3 + 1 n2(n3 + 1) - (n2 - 1) n3 + 1 (n + 1)(n - 1) n3 + 1 (n + 1)(n - 1) (n + 1)(n2 - n + 1) n - 1 n2 - n + 1 (V n + 1 0) a) Nu n = 1 th 0 1 Nu n > 1 th n - 1 < n(n - 1) + 1 < n2 - n + 1 nn khng th xy ra n - 1 n2 - n + 1 Vy gi tr ca n tm c l n = 1 b) n - 1 n2 - n + 1 n(n - 1) n2 - n + 1 (n2 - n + 1 ) - 1 n2 - n + 1 1 n2 - n + 1. C hai trng hp xy ra: + n2 - n + 1 = 1 n(n - 1) = 0 n 0

n 1 (Tm bi)

www.VNMATH.com

www.vnmath.com

12

+ n2 - n + 1 = -1 n2 - n + 2 = 0 (V nghim) Bi 3: Tm s nguyn n sao cho: a) n2 + 2n - 4 11 b) 2n3 + n2 + 7n + 1 2n - 1 c) n4 - 2n3 + 2n2 - 2n + 1 n4 - 1 d) n3 - n2 + 2n + 7 n2 + 1 Gii a) Tch n2 + 2n - 4 thnh tng hai hng t trong c mt hng t l B(11) n2 + 2n - 4 11 (n2 - 2n - 15) + 11 11 (n - 3)(n + 5) + 11 11 (n - 3)(n + 5) 11 n 3 1 1 n = B(11) + 3

n + 5 1 1 n = B(11) - 5

b) 2n3 + n2 + 7n + 1 = (n2 + n + 4) (2n - 1) + 5

2n3 + n2 + 7n + 1 2n - 1 th 5 2n - 1 hay 2n - 1 l (5) 2n 1 = - 5 n = - 22n 1 = -1 n = 02n 1 = 1 n = 12n 1 = 5 n = 3

Vy: n 2; 0; 1; 3 th 2n3 + n2 + 7n + 1 2n - 1 c) n4 - 2n3 + 2n2 - 2n + 1 n4 - 1 t A = n4 - 2n3 + 2n2 - 2n + 1 = (n4 - n3) - (n3 - n2) + (n2 - n) - (n - 1) = n3(n - 1) - n2(n - 1) + n(n - 1) - (n - 1) = (n - 1) (n3 - n2 + n - 1) = (n - 1)2(n2 + 1) B = n4 - 1 = (n - 1)(n + 1)(n2 + 1) A chia ht cho b nn n 1 A chia ht cho B n - 1 n + 1 (n + 1) - 2 n + 1

2 n + 1

n = -3n 1 = - 2 n = - 2n 1 = - 1 n = 0n 1 = 1

n 1 = 2 n = 1 (khong Tm)

Vy: n 3; 2; 0 th n4 - 2n3 + 2n2 - 2n + 1 n4 - 1 d) Chia n3 - n2 + 2n + 7 cho n2 + 1 c thng l n - 1, d n + 8 n3 - n2 + 2n + 7 n2 + 1 th n + 8 n2 + 1 (n + 8)(n - 8) n2 + 1 65 n2 + 1 Ln lt cho n2 + 1 bng 1; 5; 13; 65 ta c n bng 0; 2; 8 Th li ta c n = 0; n = 2; n = 8 (T/m) Vy: n3 - n2 + 2n + 7 n2 + 1 khi n = 0, n = 8 Bi tp v nh: Tm s nguyn n : a) n3 2 chia ht cho n 2 b) n3 3n2 3n 1 chia ht cho n2 + n + 1 c)5n 2n chia ht cho 63 Dng 4: Tn ti hay khng tn ti s chia ht Bi 1: Tm n N sao cho 2n 1 chia ht cho 7 Gii Nu n = 3k ( k N) th 2n 1 = 23k 1 = 8k - 1 chia ht cho 7 Nu n = 3k + 1 ( k N) th 2n 1 = 23k + 1 1 = 2(23k 1) + 1 = BS 7 + 1 Nu n = 3k + 2 ( k N) th 2n 1 = 23k + 2 1 = 4(23k 1) + 3 = BS 7 + 3 V y: 2n 1 chia ht cho 7 khi n = BS 3

www.VNMATH.com

www.vnmath.com

13

Bi 2: Tm n N : a) 3n 1 chia ht cho 8 b) A = 32n + 3 + 24n + 1 chia ht cho 25 c) 5n 2n chia ht cho 9 Gii a) Khi n = 2k (k N) th 3n 1 = 32k 1 = 9k 1 chia ht cho 9 1 = 8 Khi n = 2k + 1 (k N) th 3n 1 = 32k + 1 1 = 3. (9k 1 ) + 2 = BS 8 + 2 Vy : 3n 1 chia ht cho 8 khi n = 2k (k N) b) A = 32n + 3 + 24n + 1 = 27 . 32n + 2.24n = (25 + 2) 32n + 2.24n = 25. 32n + 2.32n + 2.24n = BS 25 + 2(9n + 16n) Nu n = 2k +1(k N) th 9n + 16n = 92k + 1 + 162k + 1 chia ht cho 9 + 16 = 25 Nu n = 2k (k N) th 9n c ch s tn cng bng 1 , cn 16n c ch s tn cng bng 6 suy ra 2((9n + 16n) c ch s tn cng bng 4 nn A khng chia ht cho 5 nn khng chia ht cho 25 c) Nu n = 3k (k N) th 5n 2n = 53k 23k chia ht cho 53 23 = 117 nn chia ht cho 9 Nu n = 3k + 1 th 5n 2n = 5.53k 2.23k = 5(53k 23k) + 3. 23k = BS 9 + 3. 8k = BS 9 + 3(BS 9 1)k = BS 9 + BS 9 + 3 Tng t: nu n = 3k + 2 th 5n 2n khng chia ht cho 9 www.vnmath.com

www.VNMATH.com

www.vnmath.com

14

CHUYEN E 4 TNH CHIA HET OI VI A THC A. Dang 1: Tm d cua phep chia ma khong thc hien phep chia 1. a thc chia co dang x a (a la hang) a) nh l Bdu (Bezout, 1730 1783): So d trong phep chia a thc f(x) cho nh thc x a bang gia tr cua f(x) tai x = a Ta co: f(x) = (x a). Q(x) + r ang thc ung vi moi x nen vi x = a, ta co f(a) = 0.Q(a) + r hay f(a) = r Ta suy ra: f(x) chia het cho x a f(a) = 0 b) f(x) co tong cac he so bang 0 th chia het cho x 1 c) f(x) co tong cac he so cua hang t bac chan bang tong cac he so cua cac hang t bac le th chia het cho x + 1 V du : Khong lam phep chia, hay xet xem A = x3 9x2 + 6x + 16 chia het cho B = x + 1, C = x 3 khong Ket qua: A chia het cho B, khong chia het cho C 2. a thc chia co bac hai tr len Cach 1: Tach a thc b chia thanh tong cua cac a thc chia het cho a thc chia va d Cach 2: Xet gia tr rieng: goi thng cua phep chia la Q(x), d la ax + b th f(x) = g(x). Q(x) + ax + b V du 1: Tm d cua phep chia x7 + x5 + x3 + 1 cho x2 1 Cach 1: Ta biet rang x2n 1 chia het cho x2 1 nen ta tach: x7 + x5 + x3 + 1 = (x7 x) + (x5 x) +(x3 x) + 3x + 1 = x(x6 1) + x(x4 1) + x(x2 1) + 3x + 1 chia cho x2 1 d 3x + 1 Cach 2: Goi thng cua phep chia la Q(x), d la ax + b, Ta co: x7 + x5 + x3 + 1 = (x -1)(x + 1).Q(x) + ax + b vi moi x ang thc ung vi moi x nen vi x = 1, ta co 4 = a + b (1) vi x = - 1 ta co - 2 = - a + b (2) T (1) va (2) suy ra a = 3, b =1 nen ta c d la 3x + 1 Ghi nh: an bn chia het cho a b (a -b) an + bn ( n le) chia het cho a + b (a -b) V du 2: Tm d cua cac phep chia a) x41 chia cho x2 + 1 b) x27 + x9 + x3 + x cho x2 1 c) x99 + x55 + x11 + x + 7 cho x2 + 1 Giai

www.VNMATH.com

www.vnmath.com

15

H s ca a thc chia

H s th 2 ca a thc b chia

+H s th1a thc bchia

a

a) x41 = x41 x + x = x(x40 1) + x = x[(x4)10 1] + x chia cho x4 1 d x nen chia cho x2 + 1 d x b) x27 + x9 + x3 + x = (x27 x) + (x9 x) + (x3 x) + 4x = x(x26 1) + x(x8 1) + x(x2 1) + 4x chia cho x2 1 d 4x c) x99 + x55 + x11 + x + 7 = x(x98 + 1) + x(x54 + 1) + x(x10 + 1) 2x + 7 chia cho x2 + 1 d 2x + 7 B. S o HORN 1. S o e tm ket qua cua phep chia f(x) cho x a (a la hang so), ta s dung s o horn Neu a thc b chia la a0x3 + a1x2 + a2x + a3, a thc chia la x a ta c thng la b0x2 + b1x + b2, d r th ta co

V du: a thc b chia: x3 -5x2 + 8x 4, a thc chia x 2 Ta co s o 1 - 5 8 - 4

2 1 2. 1 + (- 5) = -3 2.(- 3) + 8 = 2 r = 2. 2 +(- 4) = 0Vay: x3 -5x2 + 8x 4 = (x 2)(x2 3x + 2) + 0 la phep chia het 2. Ap dung s o Horn e tnh gia tr cua a thc tai x = a Gia tr cua f(x) tai x = a la so d cua phep chia f(x) cho x a 1. V du 1: Tnh gia tr cua A = x3 + 3x2 4 tai x = 2010 Ta co s o:

1 3 0 -4 a = 2010 1 2010.1+3 = 2013 2010.2013 + 0

= 4046130 2010.4046130 4 = 8132721296

Vay: A(2010) = 8132721296 C. Chngs minh mot a thc chia het cho mot a thc khac I. Phng phap: 1. Cach 1: Phan tch a thc b chia thanh nhan t co mot tha so la a thc chia 2. Cach 2: bien oi a thc b chia thanh mot tong cac a thc chia het cho a thc chia 3. Cach 3: Bien oi tng ng f(x) g(x) f(x) g(x) g(x) 4. cach 4: Chng to moi nghiem cua a thc chia eu la nghiem cua a thc b chia

r = ab2 + a3

a3

b2 = ab1+ a2b1= ab0+ a1

a2a1

b0 = a0

a0

a

www.VNMATH.com

www.vnmath.com

16

II. V du 1.V du 1: Chng minh rang: x8n + x4n + 1 chia het cho x2n + xn + 1 Ta co: x8n + x4n + 1 = x8n + 2x4n + 1 - x4n = (x4n + 1)2 - x4n = (x4n + x2n + 1)( x4n - x2n + 1) Ta lai co: x4n + x2n + 1 = x4n + 2x2n + 1 x2n = (x2n + xn + 1)( x2n - xn + 1) chia het cho x2n + xn + 1 Vay: x8n + x4n + 1 chia het cho x2n + xn + 1 2. V du 2: Chng minh rang: x3m + 1 + x3n + 2 + 1 chia het cho x2 + x + 1 vi moi m, n N Ta co: x3m + 1 + x3n + 2 + 1 = x3m + 1 - x + x3n + 2 x2 + x2 + x + 1 = x(x3m 1) + x2(x3n 1) + (x2 + x + 1) V x3m 1 va x3n 1 chia het cho x3 1 nen chia het cho x2 + x + 1 Vay: x3m + 1 + x3n + 2 + 1 chia het cho x2 + x + 1 vi moi m, n N 3. V du 3: Chng minh rang f(x) = x99 + x88 + x77 + ... + x11 + 1 chia het cho g(x) = x9 + x8 + x7 + ....+ x + 1 Ta co: f(x) g(x) = x99 x9 + x88 x8 + x77 x7 + ... + x11 x + 1 1 = x9(x90 1) + x8(x80 1) + ....+ x(x10 1) chia het cho x10 1 Ma x10 1 = (x 1)(x9 + x8 + x7 +...+ x + 1) chia het cho x9 + x8 + x7 +...+ x + 1 Suy ra f(x) g(x) chia het cho g(x) = x9 + x8 + x7 +...+ x + 1 Nen f(x) = x99 + x88 + x77 + ... + x11 + 1 chia het cho g(x) = x9 + x8 + x7 + ....+ x + 1 4. V du 4: CMR: f(x) = (x2 + x 1)10 + (x2 - x + 1)10 2 chia het cho g(x) = x2 x a thc g(x) = x2 x = x(x 1) co 2 nghiem la x = 0 va x = 1 Ta co f(0) = (-1)10 + 110 2 = 0 x = 0 la nghiem cua f(x) f(x) cha tha so x f(1) = (12 + 1 1)10 + (12 1 + 1)10 2 = 0 x = 1 la nghiem cua f(x) f(x) cha tha so x 1, ma cac tha so x va x 1 khong co nhan t chung, do o f(x) chia het cho x(x 1) hay f(x) = (x2 + x 1)10 + (x2 - x + 1)10 2 chia het cho g(x) = x2 x 5. V du 5: Chng minh rang a) A = x2 x9 x1945 chia het cho B = x2 x + 1 b) C = 8x9 9x8 + 1 chia het cho D = (x 1)2 c) C (x) = (x + 1)2n x2n 2x 1 chia het cho D(x) = x(x + 1)(2x + 1) Giai a) A = x2 x9 x1945 = (x2 x + 1) (x9 + 1) (x1945 x) Ta co: x2 x + 1 chia het cho B = x2 x + 1 x9 + 1 chia het cho x3 + 1 nen chia het cho B = x2 x + 1 x1945 x = x(x1944 1) chia het cho x3 + 1 (cung co nghiem la x = - 1) nen chia het cho B = x2 x + 1 Vay A = x2 x9 x1945 chia het cho B = x2 x + 1

www.VNMATH.com

www.vnmath.com

17

b) C = 8x9 9x8 + 1 = 8x9 8 - 9x8 + 9 = 8(x9 1) 9(x8 1) = 8(x 1)(x8 + x7 + ...+ 1) 9(x 1)(x7 + x6 + ...+ 1) = (x 1)(8x8 x7 x6 x5 x4 x3 x2 x 1) (8x8 x7 x6 x5 x4 x3 x2 x 1) chia het cho x 1 v co tong he so bang 0 suy ra (x 1)(8x8 x7 x6 x5 x4 x3 x2 x 1) chia het cho (x 1)2 c) a thc chia D (x) = x(x + 1)(2x + 1) co ba nghiem la x = 0, x = - 1, x = - 1

2

Ta co: C(0) = (0 + 1)2n 02n 2.0 1 = 0 x = 0 la nghiem cua C(x) C(-1) = (-1 + 1)2n (- 1)2n 2.(- 1) 1 = 0 x = - 1 la nghiem cua C(x) C(- 1

2) = (- 1

2 + 1)2n (- 1

2)2n 2.(- 1

2) 1 = 0 x = - 1

2 la nghiem cua C(x)

Moi nghiem cua a thc chia la nghiem cua a thc b chia pcm 6. V du 6: Cho f(x) la a thc co he so nguyen. Biet f(0), f(1) la cac so le. Chng minh rang f(x) khong co nghiem nguyen Gia s x = a la nghiem nguyen cua f(x) th f(x) = (x a). Q(x). Trong o Q(x) la a thc co he so nguyen, do o f(0) = - a. Q(0), f(1) = (1 a). Q(1) Do f(0) la so le nen a la so le, f(1) la so le nen 1 a la so le, ma 1 a la hieu cua 2 so le khong the la so le, mau thuan Vay f(x) khong co nghiem nguyen Bai tap ve nha: Bai 1: Tm so d khi a) x43 chia cho x2 + 1 b) x77 + x55 + x33 + x11 + x + 9 cho x2 + 1 Bai 2: Tnh gia tr cua a thc x4 + 3x3 8 tai x = 2009 Bai 3: Chng minh rang a) x50 + x10 + 1 chia het cho x20 + x10 + 1 b) x10 10x + 9 chia het cho x2 2x + 1 c) x4n + 2 + 2x2n + 1 + 1 chia het cho x2 + 2x + 1 d) (x + 1)4n + 2 + (x 1)4n + 2 chia het cho x2 + 1 e) (xn 1)(xn + 1 1) chia het cho (x + 1)(x 1)2

www.vnmath.com

www.VNMATH.com

www.vnmath.com

18

CHUYEN E 5 : SO CHNH PHNG I. So chnh phng: A. Mot so kien thc: So chnh phng: so bang bnh phng cua mot so khac V du: 4 = 22; 9 = 32 A = 4n2 + 4n + 1 = (2n + 1)2 = B2

+ S chnh phng khng tn cng bi cc ch s: 2, 3, 7, 8 + S chnh phng chia ht cho 2 th chia ht cho 4, chia ht cho 3 th chia ht cho 9, chia ht cho 5 th chia ht cho 25, chia ht cho 23 th chia ht cho 24, + S

n

11...1 = a th n

99...9 = 9a 9a + 1 = n

99...9 + 1 = 10n

B. Mot so bai toan: 1. Bai 1: Chng minh rang: Mot so chnh phng chia cho 3, cho 4 ch co the d 0 hoac 1 Giai Goi A = n2 (n N) a) xet n = 3k (k N) A = 9k2 nen chia het cho 3 n = 3k 1 (k N) A = 9k2 6k + 1, chia cho 3 d 1 Vay: so chnh phng chia cho 3 d 0 hoac 1 b) n = 2k (k N) th A = 4k2 chia het cho 4 n = 2k +1 (k N) th A = 4k2 + 4k + 1 chia cho 4 d 1 Vay: so chnh phng chia cho 4 d 0 hoac 1 Chu y: + So chnh phng chan th chia het cho 4 + So chnh phng le th chia cho 4 th d 1( Chia 8 cung d 1) 2. Bai 2: So nao trong cac so sau la so chnh phng a) M = 19922 + 19932 + 19942

b) N = 19922 + 19932 + 19942 + 19952 c) P = 1 + 9100 + 94100 + 1994100

d) Q = 12 + 22 + ...+ 1002 e) R = 13 + 23 + ... + 1003 Giai a) cac so 19932, 19942 chia cho 3 d 1, con 19922 chia het cho 3 M chia cho 3 d 2 do o M khong la so chnh phng b) N = 19922 + 19932 + 19942 + 19952 gom tong hai so chnh phng chan chia het cho 4, va hai so chnh phng le nen chia 4 d 2 suy ra N khong la so chnh phng c) P = 1 + 9100 + 94100 + 1994100 chia 4 d 2 nen khong la so chnh phng d) Q = 12 + 22 + ...+ 1002

www.VNMATH.com

www.vnmath.com

19

So Q gom 50 so chnh phng chan chia het cho 4, 50 so chnh phng le, moi so chia 4 d 1 nen tong 50 so le o chia 4 th d 2 do o Q chia 4 th d 2 nen Q khong la so chnh phng e) R = 13 + 23 + ... + 1003 Goi Ak = 1 + 2 +... + k =

k(k + 1)2

, Ak 1 = 1 + 2 +... + k = k(k - 1)

2

Ta co: Ak2 Ak -12 = k3 khi o: 13 = A12 23 = A22 A12 ..................... n3 = An2 = An - 12 Cong ve theo ve cac ang thc tren ta co:

13 + 23 + ... +n3 = An2 = 2 2

2n(n + 1) 100(100 1) 50.1012 2

la so chnh phng 3. Bai 3: CMR: Vi mi n N th cac so sau la s chnh phng. a) A = (10n +10n-1 +...+.10 +1)(10 n+1 + 5) + 1

A = (n

11.....1 )(10 n+1 + 5) + 1

1110 1.(10 5) 1

10 1

nn

t a = 10n+1 th A = a - 19

(a + 5) + 1 = 22 2a + 4a - 5 + 9 a + 4a + 4 a + 2

9 9 3

b) B = n

111.....1n - 1

555.....5 6 ( c n s 1 v n-1 s 5)

B = n

111.....1n

555.....5 + 1 = n

111.....1 . 10n +

n

555.....5 + 1 = n

111.....1 . 10n + 5

n

111.....1

+ 1

at n

11.....1 = a th 10n = 9a + 1 nen

B = a(9a + 1) + 5a + 1 = 9a2 + 6a + 1 = (3a + 1)2 = 2n - 1

33....34

c) C =2n

11.....1 .+ 44.....4 n

+ 1

at a = n

11.....1 Th C = n

11.....1n

11.....1 + 4. n

11.....1 + 1 = a. 10n + a + 4 a + 1

= a(9a + 1) + 5a + 1 = 9a2 + 6a + 1 = (3a + 1)2 d) D =

n

99....9 8n

00.....0 1 . at n

99....9 = a 10n = a + 1 D =

n

99....9 . 10n + 2 + 8. 10n + 1 + 1 = a . 100 . 10n + 80. 10n + 1

= 100a(a + 1) + 80(a + 1) + 1 = 100a2 + 180a + 81 = (10a + 9)2 = (n + 1

99....9 )2

e) E = n

11.....1n + 1

22.....2 5 = n

11.....1n + 1

22.....2 00 + 25 = n

11.....1 .10n + 2 + 2.

n

11.....100 + 25

www.VNMATH.com

www.vnmath.com

20

= [a(9a + 1) + 2a]100 + 25 = 900a2 + 300a + 25 = (30a + 5)2 = (n

33.....3 5)2

f) F = 100

44.....4 = 4.100

11.....1 la so chnh phng th 100

11.....1 la so chnh phng

So 100

11.....1 la so le nen no la so chnh phng th chia cho 4 phai d 1

That vay: (2n + 1)2 = 4n2 + 4n + 1 chia 4 d 1

100

11.....1 co hai ch so tan cung la 11 nen chia cho 4 th d 3

vay 100

11.....1 khong la so chnh phng nen F = 100

44.....4 khong la so chnh phng

Bai 4: a) Cho cc s A =

2m

11........11 ; B = m + 1

11.......11 ; C = m

66.....66

CMR: A + B + C + 8 l s chnh phng . Ta co: A

210 19

m ; B = 110 1

9

m ; C = 10 16.9

m Nen:

A + B + C + 8 = 210 19

m + 110 1

9

m + 10 16.9

m + 8 = 2 110 1 10 1 6(10 1) 72

9

m m m

= 210 1 10.10 1 6.10 6 72

9

m m m = 2 210 16.10 64 10 89 3

m m m

b) CMR: Vi mi x,y Z th A = (x+y)(x+2y)(x+3y)(x+4y) + y4 la s chnh phng. A = (x2 + 5xy + 4y2) (x2 + 5xy + 6y2) + y4 = (x2 + 5xy + 4y2) [(x2 + 5xy + 4y2) + 2y2) + y4 = (x2 + 5xy + 4y2)2 + 2(x2 + 5xy + 4y2).y2 + y4 = [(x2 + 5xy + 4y2) + y2)2 = (x2 + 5xy + 5y2)2 Bai 5: Tm so nguyen dng n e cac bieu thc sau la so chnh phng a) n2 n + 2 b) n5 n + 2 Giai a) Vi n = 1 th n2 n + 2 = 2 khong la so chnh phng Vi n = 2 th n2 n + 2 = 4 la so chnh phng Vi n > 2 th n2 n + 2 khong la so chnh phng V (n 1)2 = n2 (2n 1) < n2 (n - 2) < n2 b) Ta co n5 n chia het cho 5 V n5 n = (n2 1).n.(n2 + 1) Vi n = 5k th n chia het cho 5 Vi n = 5k 1 th n2 1 chia het cho 5 Vi n = 5k 2 th n2 + 1 chia het cho 5 Nen n5 n + 2 chia cho 5 th d 2 nen n5 n + 2 co ch so tan cung la 2 hoac 7 nen n5 n + 2 khong la so chnh phng Vay : Khong co gia tr nao cua n thoa man bai toan

www.VNMATH.com

www.vnmath.com

21

Bai 6 : a)Chng minh rang : Moi so le eu viet c di dang hieu cua hai so chnh phng b) Mot so chnh phng co ch so tan cung bang 9 th ch so hang chuc la ch so chan Giai Moi so le eu co dang a = 4k + 1 hoac a = 4k + 3 Vi a = 4k + 1 th a = 4k2 + 4k + 1 4k2 = (2k + 1)2 (2k)2 Vi a = 4k + 3 th a = (4k2 + 8k + 4) (4k2 + 4k + 1) = (2k + 2)2 (2k + 1)2 b)A la so chnh phng co ch so tan cung bang 9 nen A = (10k 3)2 =100k2 60k + 9 = 10.(10k2 6) + 9 So chuc cua A la 10k2 6 la so chan (pcm) Bai 7: Mot so chnh phng co ch so hang chuc la ch so le. Tm ch so hang n v Giai Goi n2 = (10a + b)2 = 10.(10a2 + 2ab) + b2 nen ch so hang n v can tm la ch so tan cung cua b2 Theo e bai , ch so hang chuc cua n2 la ch so le nen ch so hang chuc cua b2 phai leXet cac gia tr cua b t 0 en 9 th ch co b2 = 16, b2 = 36 co ch so hang chuc la ch so le, chung eu tan cung bang 6 Vay : n2 co ch so hang n v la 6 * Bai tap ve nha: Bai 1: Cac so sau ay, so nao la so chnh phng a) A =

50

22.....2 4 b) B = 11115556 c) C = n

99....9 n

00....0 25

d) D = n

44.....4 n - 1

88....8 9 e) M =2n

11.....1 n

22....2 f) N = 12 + 22 + ...... + 562

Bai 2: Tm so t nhien n e cac bieu thc sau la so chnh phng a) n3 n + 2 b) n4 n + 2 Bai 3: Chng minh rang a)Tong cua hai so chnh phng le khong la so chnh phng b) Mot so chnh phng co ch so tan cung bang 6 th ch so hang chuc la ch so le Bai 4: Mot so chnh phng co ch so hang chuc bang 5. Tm ch so hang n v

www.vnmath.com

www.VNMATH.com

www.vnmath.com

22

CHUYEN E 6 ONG D THC A. NH NGHA: Neu hai so nguyen a va b co cung so d trong phep chia cho mot so t nhien m 0 th ta noi a ong d vi b theo moun m, va co ong d thc: a b (mod m) V du:7 10 (mod 3) , 12 22 (mod 10) + Chu y: a b (mod m) a b m B. TNH CHAT: 1. Tnh chat phan xa: a a (mod m) 2. Tnh chat oi xng: a b (mod m) b a (mod m) 3. Tnh chat bac cau: a b (mod m), b c (mod m) th a c (mod m) 4. Cong , tr tng ve: a b (mod m) a c b d (mod m)

c d (mod m)

He qua: a) a b (mod m) a + c b + c (mod m) b) a + b c (mod m) a c - b (mod m) c) a b (mod m) a + km b (mod m) 5. Nhan tng ve : a b (mod m) ac bd (mod m)

c d (mod m)

He qua: a) a b (mod m) ac bc (mod m) (c Z) b) a b (mod m) an bn (mod m) 6. Co the nhan (chia) hai ve va moun cua mot ong d thc vi mot so nguyen dng a b (mod m) ac bc (mod mc) Chang han: 11 3 (mod 4) 22 6 (mod 8) 7. ac bc (mod m) a b (mod m)

(c, m) = 1

Chang han : 16 2 (mod 7) 8 1 (mod 7)(2, 7) = 1

C. CAC V DU: 1. V du 1: Tm so d khi chia 9294 cho 15 Giai Ta thay 92 2 (mod 15) 9294 294 (mod 15) (1) Lai co 24 1 (mod 15) (24)23. 22 4 (mod 15) hay 294 4 (mod 15) (2) T (1) va (2) suy ra 9294 4 (mod 15) tc la 9294 chia 15 th d 4 2. V du 2: Chng minh: trong cac so co dang 2n 4(n N), co vo so so chia het cho 5

www.VNMATH.com

www.vnmath.com

23

That vay: T 24 1 (mod 5) 24k 1 (mod 5) (1) Lai co 22 4 (mod 5) (2) Nhan (1) vi (2), ve theo ve ta co: 24k + 2 4 (mod 5) 24k + 2 - 4 0 (mod 5) Hay 24k + 2 - 4 chia het cho 5 vi moi k = 0, 1, 2, ... hay ta c vo so so dang 2n 4 (n N) chia het cho 5 Chu y: khi giai cac bai toan ve ong d, ta thng quan tam en a 1 (mod m) a 1 (mod m) an 1 (mod m) a -1 (mod m) an (-1)n (mod m) 3. V du 3: Chng minh rang a) 2015 1 chia het cho 11 b) 230 + 330 chi het cho 13 c) 555222 + 222555 chia het cho 7 Giai a) 25 - 1 (mod 11) (1); 10 - 1 (mod 11) 105 - 1 (mod 11) (2) T (1) va (2) suy ra 25. 105 1 (mod 11) 205 1 (mod 11) 205 1 0 (mod 11) b) 26 - 1 (mod 13) 230 - 1 (mod 13) (3) 33 1 (mod 13) 330 1 (mod 13) (4) T (3) va (4) suy ra 230 + 330 - 1 + 1 (mod 13) 230 + 330 0 (mod 13) Vay: 230 + 330 chi het cho 13 c) 555 2 (mod 7) 555222 2222 (mod 7) (5) 23 1 (mod 7) (23)74 1 (mod 7) 555222 1 (mod 7) (6) 222 - 2 (mod 7) 222555 (-2)555 (mod 7) Lai co (-2)3 - 1 (mod 7) [(-2)3]185 - 1 (mod 7) 222555 - 1 (mod 7) Ta suy ra 555222 + 222555 1 - 1 (mod 7) hay 555222 + 222555 chia het cho 7 4. V du 4: Chng minh rang so 4n + 122 + 7 chia het cho 11 vi moi so t nhien n That vay:Ta co: 25 - 1 (mod 11) 210 1 (mod 11) Xet so d khi chia 24n + 1 cho 10. Ta co: 24 1 (mod 5) 24n 1 (mod 5) 2.24n 2 (mod 10) 24n + 1 2 (mod 10) 24n + 1 = 10 k + 2 Nen 4n + 122 + 7 = 210k + 2 + 7 =4. 210k + 7 = 4.(BS 11 + 1)k + 7 = 4.(BS 11 + 1k) + 7 = BS 11 + 11 chia het cho 11 Bai tap ve nha: Bai 1: CMR: a) 228 1 chia het cho 29 b)Trong cac so co dang2n 3 co vo so so chia het cho 13 Bai 2: Tm so d khi chia A = 2011 + 2212 + 19962009 cho 7.

www.vnmath.com

www.VNMATH.com

www.vnmath.com

24

CHUYEN E 7 CAC BAI TOAN VE BIEU THC HU T A. Nhac lai kien thc: Cac bc rut gon bieu thc hu t a) Tm KX: Phan tch mau thanh nhan t, cho tat ca cac nhan t khac 0 b) Phan tch t thanh nhan , chia t va mau cho nhan t chung B. Bai tap:

Bai 1: Cho bieu thc A = 4 2

4 2

5 410 9

x xx x

a) Rut gon A b) tm x e A = 0 c) Tm gia tr cua A khi 2 1 7x Giai a)kx : x4 10x2 + 9 0 [(x2)2 x2] (9x2 9) 0 x2(x2 1) 9(x2 1) 0 (x2 1)(x2 9) 0 (x 1)(x + 1)(x 3)(x + 3) 0 1

3

xx

T : x4 5x2 + 4 = [(x2)2 x2] (x2 4) = x2(x2 1) 4(x2 1) = (x2 1)(x2 4) = (x 1)(x + 1)(x 2)(x + 2)

Vi x 1; x 3 th A = (x - 1)(x + 1)(x - 2)(x + 2) (x - 2)(x + 2)(x - 1)(x + 1)(x - 3)(x + 3) (x - 3)(x + 3)

b) A = 0 (x - 2)(x + 2)(x - 3)(x + 3)

= 0 (x 2)(x + 2) = 0 x = 2

c) 2 1 7x 2 1 7 2 8 42 1 7 2 6 3

x x xx x x

* Vi x = 4 th A = (x - 2)(x + 2) (4 - 2)(4 + 2) 12(x - 3)(x + 3) (4 - 3)(4 + 3) 7

* Vi x = - 3 th A khong xac nh 2. Bai 2:

Cho bieu thc B = 3 2

3 2

2x 7x 12x 453x 19x 33x 9

a) Rut gon B b) Tm x e B > 0 Giai a) Phan tch mau: 3x3 19x2 + 33x 9 = (3x3 9x2) (10x2 30x) + (3x 9) = (x 3)(3x2 10x + 3) = (x 3)[(3x2 9x) (x 3)] = (x 3)2(3x 1) kx: (x 3)2(3x 1) 0 x 3 va x 1

3

b) Phan tch t, ta co: 2x3 7x2 12x + 45 = (2x3 6x2 ) - (x2 - 3x) (15x - 45) = (x 3)(2x2 x 15)

www.VNMATH.com

www.vnmath.com

25

= (x 3)[(2x2 6x) + (5x 15)] = (x 3)2(2x + 5) Vi x 3 va x 1

3

Th B = 3 2

3 2

2 7 12 453 19 33 9

x x xx x x =

2

2

(x - 3) (2x + 5) 2x + 5(x - 3) (3x - 1) 3x - 1

c) B > 0 2x + 53x - 1

> 0

13

3 1 0 5 12 5 0 2 3

53 1 0 1232 5 0

52

x

xx xx

x xxx

x

3. Bai 3 Cho bieu thc C = 2 2

1 2 5 1 2:1 1 1 1

x xx x x x

a) Rut gon bieu thc C b) Tm gia tr nguyen cua x e gia tr cua bieu thc B la so nguyen Giai a) kx: x 1 C = 2 2

1 2 5 1 2 1 2(1 ) 5 ( 1)( 1) 2: .1 1 1 1 (1 )(1 ) 1 2 2 1

x x x x x xx x x x x x x x

b) B co gia tr nguyen khi x la so nguyen th 22 1x co gia tr nguyen

2x 1 la (2) 2 1 1 12 1 1 02 1 2 1,52 1 2 1

x xx xx xx x

oi chieu kx th ch co x = 0 thoa man 4. Bai 4

Cho bieu thc D = 3 2

2

22 4

x x xx x x

a) Rut gon bieu thc D b) Tm x nguyen e D co gia tr nguyen c) Tm gia tr cua D khi x = 6 Giai a) Neu x + 2 > 0 th 2x = x + 2 nen D =

3 2

2

22 4

x x xx x x

=

3 2 2

2

2 ( 1)( 2)( 2) 4 ( 2) ( 2)( 2) 2x x x x x x x x

x x x x x x x

Neu x + 2 < 0 th 2x = - (x + 2) nen

www.VNMATH.com

www.vnmath.com

26

D = 3 2

2

22 4

x x xx x x

=

3 2

2

2 ( 1)( 2)( 2) 4 ( 2) ( 2)( 2) 2x x x x x x x

x x x x x x x

Neu x + 2 = 0 x = -2 th bieu thc D khong xac nh b) e D co gia tr nguyen th

2

2x x hoac

2x co gia tr nguyen

+) 2

2x x co gia tr nguyen

2 x(x - 1) 2 x - x 2 x > - 2x > - 2

V x(x 1) la tch cua hai so nguyen lien tiep nen chia het cho 2 vi moi x > - 2

+) 2x co gia tr nguyen x 2 x = 2k 2k (k Z; k < - 1)

x < - 2 x < - 2x

c) Khia x = 6 x > - 2 nen D = 2

2x x = 6(6 1) 15

2

* Dang 2: Cac bieu thc co tnh quy luat Bai 1: Rut gon cac bieu thc a) A = 22 2

3 5 2 1......(1.2) (2.3) ( 1)

nn n

Phng phap: Xuat phat t hang t cuoi e tm ra quy luat Ta co 2

2 1( 1)

nn n

= 2 2 2 2

2 1 1 1( 1) ( 1)n

n n n n Nen

A = 2 2 2 2 2 2 2 2 2 21 1 1 1 1 1 1 1 1 1 ( 1)......1 2 2 3 3 ( 1) 1 ( 1) ( 1)

n nn n n n n

b) B = 2 2 2 21 1 1 11 . 1 . 1 ........ 12 3 4 n

Ta co 2

2 2 2

1 1 ( 1)( 1)1 k k kk k k

Nen

B = 2 2 2 2 2 2 2 21.3 2.4 3.5 ( 1)( 1) 1.3.2.4...( 1)( 1) 1.2.3...( 1) 3.4.5...( 1) 1 1 1. . ... . .2 3 4 2 .3 .4 ... 2.3.4...( 1) 2.3.4.... 2 2

n n n n n n n nn n n n n n n

c) C = 150 150 150 150......5.8 8.11 11.14 47.50

= 1 1 1 1 1 1 1150. . ......3 5 8 8 11 47 50

= 50. 1 1 950. 455 50 10

d) D = 1 1 1 1......1.2.3 2.3.4 3.4.5 ( 1) ( 1)n n n

=1 1 1 1 1 1 1. ......2 1.2 2.3 2.3 3.4 ( 1) ( 1)n n n n

= 1 1 1 ( 1)( 2)2 1.2 ( 1) 4 ( 1)

n nn n n n

Bai 2: a) Cho A = 1 2 2 1...

1 2 2 1m m

m n ; B =

1 1 1 1......2 3 4 n . Tnh A

B

Ta co

www.VNMATH.com

www.vnmath.com

27

A = 1

1 1 1 1... 1 1 ... 1 ... ( 1)1 2 2 1 1 2 2 1n

n n n n n nn n n n

= 1 1 1 1 1 1 1... 1 ... nB1 2 2 1 2 2 1

n nn n n n

AB

= n

b) A = 1 1 1 1......1.(2n - 1) 3.(2n - 3) (2n - 3).3 (2n - 1).1

; B = 1 + 1 1......3 2n - 1

Tnh A : B Giai

A = 1 1 1 1 1 1 11 ... 12n 2n - 1 3 2n - 3 2n - 3 3 2n - 1

1 1 1 1 1 1 11 ...... ...... 12n 3 2n - 1 2n - 3 2n - 1 2n - 3 31 1 1 1 1 A 1.2. 1 ...... .2.B

2n 3 2n - 1 2n - 3 2n B n

Bai tap ve nha Rut gon cac bieu thc sau:

a) 1 1 1+......+1.2 2.3 (n - 1)n

b) 2 2 2 2

2 2 2 2

1 3 5 n. . ......2 1 4 1 6 1 (n + 1) 1

c) 1 1 1+......+1.2.3 2.3.4 n(n + 1)(n +2)

* Dang 3: Rut gon; tnh gia tr bieu thc thoa man ieu kien cua bien

Bai 1: Cho 1x 3x

+ = . Tnh gi tr ca cc biu thc sau :

a) 22

1A x

x= + ; b) 3

3

1B x

x= + ; c) 4

4

1C x

x= + ; d) 5

5

1D x

x= + .

Li gii

a) 2

22

1 1A x x 2 9 2 7

x x

= + = + - = - = ;

b) 3

33

1 1 1B x x 3 x 27 9 18

x x x

= + = + - + = - = ;

c) 2

4 24 2

1 1C x x 2 49 2 47

x x

= + = + - = - = ;

d) 2 3 52 3 5

1 1 1 1A.B x x x x D 3

x x x x

= + + = + + + = + D = 7.18 3 = 123.

Bai 2: Cho x y z + + = 2

a b c (1); a b c+ + = 2

x y z (2).

Tnh gia tr bieu thc D = 22 2a b c + +

x y z

www.VNMATH.com

www.vnmath.com

28

T (1) suy ra bcx + acy + abz = 0 (3) T (2) suy ra

2 22 2 2 2a b c ab ac bc a b c ab ac bc + + + 2 . 4 + + 4 2 .x y z xy xz yz x y z xy xz yz

(4) Thay (3) vao (4) ta co D = 4 2.0 = 4 Bai 3 a) Cho abc = 2; rut gon bieu thc A = a b 2c

ab + a + 2 bc + b + 1 ac + 2c + 2

Ta co : A = a ab 2c a ab 2cab + a + 2 abc + ab + a ac + 2c + 2 ab + a + 2 2 + ab + a ac + 2c + abc

= a ab 2c a ab 2 ab + a + 2 1ab + a + 2 2 + ab + a c(a + 2 + ab) ab + a + 2 2 + ab + a a + 2 + ab ab + a + 2

b) Cho a + b + c = 0; rut gon bieu thc B = 2 2 2

2 2 2 2 2 2 2 2 2

a b ca - b - c b - c - a c - b - a

T a + b + c = 0 a = -(b + c) a2 = b2 + c2 + 2bc a2 - b2 - c2 = 2bc Tng t ta co: b2 - a2 - c2 = 2ac ; c2 - b2 - a2 = 2ab (Hoan v vong quanh), nen

B = 2 2 2 3 3 3a b c a b c

2bc 2ac 2ab 2abc (1)

a + b + c = 0 -a = (b + c) -a3 = b3 + c3 + 3bc(b + c) -a3 = b3 + c3 3abc a3 + b3 + c3 = 3abc (2) Thay (2) vao (1) ta co B =

3 3 3a b c 3abc 32abc 2abc 2 (V abc 0)

c) Cho a, b, c tng oi mot khac nhau thoa man: (a + b + c)2 = a2 + b2 + c2

Rut gon bieu thc C = 2 2 2

2 2 2

a b c + a + 2bc b + 2ac c + 2ab

T (a + b + c)2 = a2 + b2 + c2 ab + ac + bc = 0 a2 + 2bc = a2 + 2bc (ab + ac + bc) = a2 ab + bc ac = (a b)(a c) Tng t: b2 + 2 ac = (b a)(b c) ; c2 + 2ab = (c a)(c b)

C = 2 2 2 2 2 2a b c a b c + -

(a - b)(a - c) (b - a)(b - c) (c - a)(c - b) (a - b)(a - c) (a - b)(b - c) (a - c)(b - c)

= 2 2 2a (b - c) b (a - c) c (b - c) (a - b)(a - c)(b - c) - 1

(a - b)(a - c)(b - c) (a - b)(a - c)(b - c) (a - b)(a - c)(b - c) (a - b)(a - c)(b - c)

* Dang 4: Chng minh ang thc thoa man ieu kien cua bien 1. Bai 1: Cho 1 1 1 + + = 2

a b c (1); 2 2 2

1 1 1+ + = 2a b c

(2).

Chng minh rang: a + b + c = abc T (1) suy ra

2 2 2 2 2 2

1 1 1 1 1 1 1 1 1 1 1 1 + + + 2. + + 4 2. + + 4 + + a b c ab bc ac ab bc ac a b c

1 1 1 a + b + c + + 1 1ab bc ac abc

a + b + c = abc

www.VNMATH.com

www.vnmath.com

29

2. Bai 2: Cho a, b, c 0 v a + b + c 0 tha mn iu kin 1 1 1 1a b c a b c

+ + =+ +

.

Chng minh rng trong ba s a, b, c c hai s i nhau.

T suy ra rng :2009 2009 2009 2009 2009 2009

1 1 1 1

a b c a b c+ + =

+ +.

Ta c : 1 1 1 1a b c a b c

+ + =+ +

1 1 1 1 0a b c a b c

+ + - =+ +

a b a b 0ab c(a b c)

+ ++ =+ +

a b 0 a b

c(a b c) ab(a b). 0 (a + b)(b + c)(c + a) = 0 b c 0 b c

abc(a b c)c a 0 c a

+ = = - + + + + = + = = - + + + = = -

T suy ra : 2009 2009 2009 2009 2009 2009 2009

1 1 1 1 1 1 1

a b c a ( c) c a+ + = + + =

-

2009 2009 2009 2009 2009 2009 2009

1 1 1

a b c a ( c) c a= =

+ + + - +

2009 2009 2009 2009 2009 2009

1 1 1 1

a b c a b c+ + =

+ +.

3. Bai 3: Cho a b c b c a + + b c a a b c

(1) chng minh rang : trong ba so a, b, c ton tai hai so bang nhau T (1) 2 2 2 2 2 2 2 2 2a c + ab + bc = b c + ac + a b a (b - c) - a(c b ) bc(c - b) = 0 (c b)(a2 ac = ab + bc) = 0 (c b)(a b)( a c) = 0 pcm 4. Bai 4: Cho (a2 bc)(b abc) = (b2 ac)(a abc); abc 0 va a b Chng minh rang: 1 1 1 + + = a + b + c

a b c

T GT a2b b2c - a3bc + ab2c2 = ab2 a2c ab3c + a2bc2 (a2b ab2) + (a2c b2c) = abc2(a b) + abc(a - b)(a + b) (a b)(ab + ac + bc) = abc(a b)(a + b + c) ab + ac + bc = a + b + c

abc 1 1 1+ + = a + b + c

a b c

5. Bai 5: Cho a + b + c = x + y + z = a b c + + = 0

x y z. Chng minh rang: ax2 + by2 + cz2 = 0

T x + y + z = 0 x2 = (y + z)2 ; y2 = (x + z)2 ; z2 = (y + x)2 ax2 + by2 + cz2 = a(y + z)2 + b(x + z)2 + c (y + x)2 = = (b + c)x2 + (a + c)y2 + (a + b)z2 + 2(ayz + bxz + cxy) (1) T a + b + c = 0 - a = b + c; - b = a + c; - c = a + b (2) T a b c + + = 0

x y z ayz + bxz + cxy = 0 (3). Thay (2), (3) vao (1); ta co:

ax2 + by2 + cz2 = -( ax2 + by2 + cz2 ) ax2 + by2 + cz2 = 0 6. Bai 6:

www.VNMATH.com

www.vnmath.com

30

Cho a b c + 0b - c c - a a - b

; chng minh: 2 2 2a b c+ 0(b - c) (c - a) (a - b)

T a b c + 0b - c c - a a - b

2 2a b c b ab + ac - c =

b - c a - c b - a (a - b)(c - a)

2 2

2

a b ab + ac - c(b - c) (a - b)(c - a)(b - c)

(1) (Nhan hai ve vi 1b - c

)

Tng t, ta co: 2 2

2

b c bc + ba - a(c - a) (a - b)(c - a)(b - c)

(2) ; 2 2

2

c a ac + cb - b(a - b) (a - b)(c - a)(b - c)

(3) Cong tng ve (1), (2) va (3) ta co pcm 7. Bai 7: Cho a + b + c = 0; chng minh: a - b b - c c - a c a b + +

c a b a - b b - c c - a = 9 (1)

at a - b b - c c - a = x ; ; c a b

y z c 1 a 1 b 1 = ; a - b x b - c c - a y z

(1) 1 1 1x + y + z + + 9x y z

Ta co: 1 1 1 y + z x + z x + yx + y + z + + 3 + + x y z x y z

(2)

Ta lai co: 2 2y + z b - c c - a c b bc + ac - a c c(a - b)(c - a - b) c(c - a - b). .

x a b a - b ab a - b ab(a - b) ab

= 2c 2c - (a + b + c) 2cab ab

(3)

Tng t, ta co: 2x + z 2a

y bc (4) ;

2x + y 2bz ac

(5) Thay (3), (4) va (5) vao (2) ta co:

1 1 1x + y + z + + 3x y z

+ 2 2 2 2c 2a 2b

ab bc ac = 3 + 2

abc(a3 + b3 + c3 ) (6)

T a + b + c = 0 a3 + b3 + c3 = 3abc (7) ? Thay (7) vao (6) ta co: 1 1 1x + y + z + + 3

x y z +

2 abc

. 3abc = 3 + 6 = 9

Bai tap ve nha: Bai 1:

Cho bieu thc A = 22 3 2 : 1

3 2 5 6 1x x x x

x x x x x

a) Rut gon A b) Tm x e A = 0; A > 0 Bai 2:

Cho bieu thc B = 3 2

3 2

3 7 5 12 4 3y y yy y y

a) Rut gon B

www.VNMATH.com

www.vnmath.com

31

b) Tm so nguyen y e 2D2y + 3

co gia tr nguyen

c) Tm so nguyen y e B 1 Bai 3 : cho 1 1 1 + + 0

x y z ; tnh gia tr bieu thc A = 2 2 2yz xz xy+ + x y z

HD: A = 3 3 3xyz xyz xyz + + x y z

; van dung a + b + c = 0 a3 + b3 + c3 = 3abc Bai 4:

Cho a3 + b3 + c3 = 3abc ; Tnh gia tr bieu thc A = a b c+ 1 + 1 + 1b c a

Bai 5: Cho x + y + z = 0; chng minh rang: 3 0y z x z x y

x y z

Bai 6: Cho a + b + c = a2 + b2 + c2 = 1; a b c

x y z . Chng minh xy + yz + xz = 0

www.vnmath.com

www.VNMATH.com

www.vnmath.com

32

CHUYEN E 8 - CAC BAI TOAN VE NH L TA-LET A.Kien thc: 1. nh l Ta-let:

* nh l Talet ABCMN // BC

AM AN = AB AC

* He qua: MN // BC AM AN MN =

AB AC BC

B. Bai tap ap dung: 1. Bai 1: Cho t giac ABCD, ng thang qua A song song vi BC cat BD E, ng thang qua B song song vi AD cat AC G a) chng minh: EG // CD b) Gia s AB // CD, chng minh rang AB2 = CD. EG Giai Goi O la giao iem cua AC va BD a) V AE // BC OE OA =

OB OC (1)

BG // AC OB OG = OD OA

(2)

Nhan (1) vi (2) ve theo ve ta co: OE OG = OD OC

EG // CD b) Khi AB // CD th EG // AB // CD, BG // AD nen

2AB OA OD CD AB CD = = AB CD. EGEG OG OB AB EG AB

Bai 2: Cho ABC vuong tai A, Ve ra pha ngoai tam giac o cac tam giac ABD vuong can B, ACF vuong can C. Goi H la giao iem cua AB va CD, K la giao iem cua AC va BF. Chng minh rang: a) AH = AK b) AH2 = BH. CK Giai at AB = c, AC = b. BD // AC (cung vuong goc vi AB) nen AH AC b AH b AH b

HB BD c HB c HB + AH b + c

Hay AH b AH b b.cAHAB b + c c b + c b + c

(1)

NM

CB

A

HFK

D

CB

A

O

GE

D C

B

A

www.VNMATH.com

www.vnmath.com

33

AB // CF (cung vuong goc vi AC) nen AK AB c AK c AK cKC CF b KC b KC + AK b + c

Hay AK b AK c b.cAKAC b + c b b + c b + c

(2) T (1) va (2) suy ra: AH = AK b) T AH AC b

HB BD c va AK AB c

KC CF b suy ra AH KC AH KC

HB AK HB AH (V AH = AK)

AH2 = BH . KC 3. Bai 3: Cho hnh bnh hanh ABCD, ng thang a i qua A lan lt cat BD, BC, DC theo th t tai E, K, G. Chng minh rang: a) AE2 = EK. EG b) 1 1 1

AE AK AG

c) Khi ng thang a thay oi v tr nhng van qua A th tch BK. DG co gia tr khong oi Giai a) V ABCD la hnh bnh hanh va K BC nen AD // BK, theo he qua cua nh l Ta-let ta co:

2EK EB AE EK AE = = AE EK.EGAE ED EG AE EG

b) Ta co: AE DE = AK DB

; AE BE = AG BD

nen

AE AE BE DE BD 1 1 = 1 AE 1AK AG BD DB BD AK AG

1 1 1

AE AK AG (pcm)

c) Ta co: BK AB BK a = = KC CG KC CG

(1); KC CG KC CG = = AD DG b DG

(2)

Nhan (1) vi (2) ve theo ve ta co: BK a= BK. DG = abb DG

khong oi (V a = AB; b = AD la o dai hai canh cua hnh bnh hanh ABCD khong oi) 4. Bai 4: Cho t giac ABCD, cac iem E, F, G, H theo th t chia trong cac canh AB, BC, CD, DA theo t so 1:2. Chng minh rang: a) EG = FH b) EG vuong goc vi FH Giai Goi M, N theo th t la trung iem cua CF, DG Ta co CM = 1

2 CF = 1

3BC BM 1=

BC 3 BE BM 1= =

BA BC 3

EM // AC EM BM 2 2 = EM = ACAC BE 3 3

(1)

Tng t, ta c: NF // BD NF CF 2 2 = NF = BDBD CB 3 3

(2)

G

b

a

EK

D C

BA

Q

PO

N M

H F

G

E

D

C

B

A

www.VNMATH.com

www.vnmath.com

34

m AC = BD (3) T (1), (2), (3) suy ra : EM = NF (a)

Tng t nh trn ta c: MG // BD, NH // AC v MG = NH = 13

AC (b)

Mt khc EM // AC; MG // BD V AC BD EM MG 0EMG = 90 (4) Tng t, ta c: 0FNH = 90 (5) T (4) v (5) suy ra 0EMG = FNH = 90 (c) T (a), (b), (c) suy ra EMG = FNH (c.g.c) EG = FH b) Gi giao im ca EG v FH l O; ca EM v FH l P; ca EM v FN l Q th 0PQF = 90 0QPF + QFP = 90 m QPF = OPE (i nh), OEP = QFP (EMG = FNH) Suy ra 0EOP = PQF = 90 EO OP EG FH 5. Bi 5: Cho hnh thang ABCD c y nh CD. T D v ng thng song song vi BC, ct AC ti M v AB ti K, T C v ng thng song song vi AD, ct AB ti F, qua F ta li v ng thng song song vi AC, ct BC ti P. Chng minh rng a) MP // AB b) Ba ng thng MP, CF, DB ng quy Gii

a) EP // AC CP AF = PB FB

(1)

AK // CD CM DC = AM AK

(2)

cc t gic AFCD, DCBK la cc hnh bnh hnh nn AF = DC, FB = AK (3)

Kt hp (1), (2) v (3) ta c CP CMPB AM

MP // AB (nh l Ta-lt o) (4)

b) Gi I l giao im ca BD v CF, ta c: CP CMPB AM

= DC DCAK FB

M DC DIFB IB

(Do FB // DC) CP DIPB IB

IP // DC // AB (5) T (4) v (5) suy ra : qua P c hai ng thng IP, PM cng song song vi AB // DC nn theo tin clt th ba im P, I, M thng hang hay MP i qua giao im ca CF v DB hay ba ng thng MP, CF, DB ng quy 6. Bi 6: Cho ABC c BC < BA. Qua C k ng thng vung goc vi tia phn gic BE ca ABC ; ng thng ny ct BE ti F v ct trung tuyn BD ti G. Chng minh rng on thng EG b on thng DF chia lm hai phn bng nhau Gii Gi K l giao im ca CF v AB; M l giao im ca DF v BC KBC c BF va l phn gic va l ng cao nn KBC cn ti B BK = BC v FC = FK

I P

FK

M

D C

BA

www.VNMATH.com

www.vnmath.com

35

Mt khc D l trung im AC nn DF l ng trung bnh ca AKC DF // AK hay DM // AB Suy ra M l trung im ca BC DF = 1

2AK (DF l ng trung bnh ca AKC), ta c

BG BK = GD DF

( do DF // BK) BG BK 2BK = GD DF AK

(1)

Mt khc CE DC - DE DC AD1 1DE DE DE DE

(V AD = DC)

CE AE - DE DC AD1 1DE DE DE DE

Hay CE AE - DE AE AB1 2 2DE DE DE DF

(v AEDE

= ABDF

: Do DF // AB)

Suy ra CE AK + BK 2(AK + BK)2 2DE DE AK

(Do DF = 12

AK) CE 2(AK + BK) 2BK2DE AK AK

(2)

T (1) v (2) suy ra BGGD

= CEDE

EG // BC

Gi giao im ca EG v DF l O ta c OG OE FO = = MC MB FM

OG = OE Bi tp v nh Bi 1: Cho t gic ABCD, AC v BD ct nhau ti O. ng thng qua O v song song vi BC ct AB E; ng thng song song vi CD qua O ct AD ti F a) Chng minh FE // BD b) T O k cc ng thng song song vi AB, AD ct BD, CD ti G v H. Chng minh: CG. DH = BG. CH Bi 2: Cho hnh bnh hnh ABCD, im M thuc cnh BC, im N thuc tia i ca tia BC sao cho BN = CM; cc ng thng DN, DM ct AB theo th t ti E, F. Chng minh: a) AE2 = EB. FE

b) EB =2AN

DF . EF

M

GK

F

D E C

B

A

www.VNMATH.com

www.vnmath.com

36

CHUYEN E 9 CAC BAI TOAN S DUNG NH L TALET VA TNH CHAT NG PHAN GIAC

A. Kien thc: 1. nh l Ta-let:

* nh l Talet ABCMN // BC

AM AN = AB AC

* He qua: MN // BC AM AN MN = AB AC BC

2. Tnh chat ng phan giac: ABC ,AD la phan giac goc A BD AB =

CD AC

ADla phan giac goc ngoai tai A: BD' AB = CD' AC

B. Bai tap van dung 1. Bai 1: Cho ABC co BC = a, AB = b, AC = c, phan giac AD a) Tnh o dai BD, CD b) Tia phan giac BI cua goc B cat AD I; tnh t so: AI

ID

Giai a) AD la phan giac cua BAC nen BD AB c

CD AC b

BD c BD c acBD = CD + BD b + c a b + c b + c

Do o CD = a - acb + c

= abb + c

b) BI la phan giac cua ABC nen AI AB ac b + cc : ID BD b + c a

2. Bai 2: Cho ABC, co B < 600 phan giac AD a) Chng minh AD < AB b) Goi AM la phan giac cua ADC. Chng minh rang BC > 4 DM Giai

a)Ta co AADB = C + 2

> A + C

2 =

00180 - B 60

2

ADB > B AD < AB b) Goi BC = a, AC = b, AB = c, AD = d Trong ADC, AM la phan giac ta co DM AD = CM AC

DM AD DM AD = = CM + DM AD + AC CD AD + AC

D' CB

AD CB

A

a

cb

I

D CB

A

M D BC

A

NM

CB

A

www.VNMATH.com

www.vnmath.com

37

DM = CD.AD CD. dAD + AC b + d

; CD = abb + c

( Van dung bai 1) DM = abd(b + c)(b + d)

e c/m BC > 4 DM ta c/m a > 4abd(b + c)(b + d)

hay (b + d)(b + c) > 4bd (1)

That vay : do c > d (b + d)(b + c) > (b + d)2 4bd . Bat ang thc (1) c c/m Bai 3: Cho ABC, trung tuyen AM, cac tia phan giac cua cac goc AMB , AMC cat AB, AC theo th t D va E a) Chng minh DE // BC b) Cho BC = a, AM = m. Tnh o dai DE c) Tm tap hp cac giao diem I cua AM va DE neu ABC co BC co nh, AM = m khong oi d) ABC co ieu kien g th DE la ng trung bnh cua no Giai a) MD la phan giac cua AMB nen DA MB

DB MA (1)

ME la phan giac cua AMC nen EA MCEC MA

(2)

T (1), (2) va gia thiet MB = MC ta suy ra DA EADB EC

DE // BC

b) DE // BC DE AD AIBC AB AM

. at DE = x xm - x 2a.m2 x =

a m a + 2m

c) Ta co: MI = 12

DE = a.ma + 2m

khong oi I luon cach M mot oan khong oi nen

tap hp cac iem I la ng tron tam M, ban knh MI = a.ma + 2m

(Tr giao iem cua no

vi BC d) DE la ng trung bnh cua ABC DA = DB MA = MB ABC vuong A4. Bai 4: Cho ABC ( AB < AC) cac phan giac BD, CE a) ng thang qua D va song song vi BC cat AB K, chng minh E nam gia B va K b) Chng minh: CD > DE > BE Giai a) BD la phan giac nen AD AB AC AE AD AE = < = DC BC BC EB DC EB

(1)

Mat khac KD // BC nen AD AKDC KB

(2)

T (1) va (2) suy ra AK AE AK + KB AE + EBKB EB KB EB

AB AB KB > EBKB EB

ED

M

I

CB

A

E

D

M

K

CB

A

www.VNMATH.com

www.vnmath.com

38

E nam gia K va B b) Goi M la giao iem cua DE va CB. Ta co CBD = KDB (so le trong) KBD = KDB ma E nam gia K va B nen KDB > EDB KBD > EDB EBD > EDB EB < DE Ta lai co CBD + ECB = EDB + DEC DEC > ECB DEC > DCE (V DCE = ECB ) Suy ra: CD > ED CD > ED > BE 5. Bai 5: Cho ABC . Ba ng phan giac AD, BE, CF. Chng minh

a. DB EC FA. . 1DC EA FB

. b. 1 1 1 1 1 1

AD BE CF BC CA AB .

Giai a)AD la ng phan giac cua BAC nen ta co: DB AB =

DC AC (1)

Tng t: vi cac phan giac BE, CF ta co: EC BC = EA BA

(2) ; FA CA = FB CB

(3)

T (1); (2); (3) suy ra: DB EC FA AB BC CA. . = . .DC EA FB AC BA CB

= 1

b) t AB = c , AC = b , BC = a , AD = da. Qua C k ng thng song song vi AD , ct tia BA H. Theo L Talt ta c: AD BA

CH BH BA.CH c.CH cAD .CH

BH BA + AH b + c

Do CH < AC + AH = 2b nn: 2abcd

b c

1 1 1 1 1 1 1 12 2 2a a

b cd bc b c d b c

Chng minh tng t ta c : 1 1 1 12bd a c V

1 1 1 12cd a b Nn:

1 1 1 1 1 1 1 1 1 12a b cd d d b c a c a b

1 1 1 1 1 1 1.22a b cd d d a b c

1 1 1 1 1 1

a b cd d d a b c ( pcm ) Bi tp v nh Cho ABC co BC = a, AC = b, AB = c (b > c), cac phan giac BD, CE a) Tnh o dai CD, BE roi suy ra CD > BE b) Ve hnh bnh hanh BEKD. Chng minh: CE > EK c) Chng minh CE > BD

www.vnmath.com

H

FE

DCB

A

www.VNMATH.com

www.vnmath.com

39

CHUYEN E 10 CAC BAI TOAN VE TAM GIAC ONG DANG A. Kien thc: * Tam giac ong dang: a) trng hp th nhat: (c.c.c) ABC ABC AB AC BC = =

A'B' A'C' B'C'

b) trng hp th nhat: (c.g.c) ABC ABC AB AC =

A'B' A'C' ; A = A'

c. Trng hp ong dang th ba (g.g) ABC ABC A = A' ; B = B' AH; AHla hai ng cao tng ng th: A'H'

AH = k (T so ong dang); A'B'C'

ABC

SS

= K2

B. Bai tap ap dung Bai 1: Cho ABC co B = 2 C , AB = 8 cm, BC = 10 cm. a)Tnh AC b)Neu ba canh cua tam giac tren la ba so t nhien lien tiep th moi canh la bao nhieu? Giai Cach 1: Tren tia oi cua tia BA lay iem E sao cho:BD = BC ACD ABC (g.g) AC AD

AB AC

2AC AB. AD =AB.(AB + BD) = AB(AB + BC) = 8(10 + 8) = 144 AC = 12 cm Cach 2: Ve tia phan giac BE cua ABC ABE ACB

2AB AE BE AE + BE AC = AC = AB(AB + CB) AC AB CB AB + CB AB + CB

= 8(8 + 10) = 144 AC = 12 cm b) Goi AC = b, AB = a, BC = c th t cau a ta co b2 = a(a + c) (1) V b > anen co the b = a + 1 hoac b = a + 2 + Neu b = a + 1 th (a + 1)2 = a2 + ac 2a + 1 = ac a(c 2) = 1 a = 1; b = 2; c = 3(loai) + Neu b = a + 2 th a(c 4) = 4 - Vi a = 1 th c = 8 (loai) - Vi a = 2 th c = 6 (loai)

E

D

C

B

A

D

CB

A

www.VNMATH.com

www.vnmath.com

40

- vi a = 4 th c = 6 ; b = 5 Vay a = 4; b = 5; c = 6 Bai 2: Cho ABC can tai A, ng phan giac BD; tnh BD biet BC = 5 cm; AC = 20 cm Giai Ta co CD BC 1 =

AD AC 4 CD = 4 cm va BC = 5 cm

Bai toan tr ve bai 1 Bai 3: Cho ABC can tai A va O la trung iem cua BC. Mot iem O di ong tren AB, lay iem E tren AC sao cho

2OBCE = BD

. Chng minh rang

a) DBO OCE b) DOE DBO OCE c) DO, EO lan lt la phan giac cua cac goc BDE, CED d) khoang cach t O en oan ED khong oi khi D di ong tren AB Giai

a) T 2OBCE =

BD CE OB =

OB BD va B = C (gt) DBO OCE

b) T cau a suy ra 23O = E (1) V B, O ,C thang hang nen 03O + DOE EOC 180 (2) trong tam giac EOC th 02E + C EOC 180 (3) T (1), (2), (3) suy ra DOE B C DOE va DBO co DO OE =

DB OC (Do DBO OCE)

va DO OE = DB OB

(Do OC = OB) va DOE B C nen DOE DBO OCE c) T cau b suy ra 1 2D = D DO la phan giac cua cac goc BDE Cung t cau b suy ra 1 2E = E EO la phan giac cua cac goc CED c) Goi OH, OI la khoang cach t O en DE, CE th OH = OI, ma O co nh nen OH khong oi OI khong oi khi D di ong tren AB Bai 4: (e HSG huyen Loc ha nam 2007 2008) Cho ABC can tai A, co BC = 2a, M la trung iem BC, lay D, E thuoc AB, AC sao cho DME = B a) Chng minh tch BD. CE khong oi b)Chng minh DM la tia phan giac cua BDE c) Tnh chu vi cua AED neu ABC la tam giac eu Giai

21

3

21 H

I

O

E

D

CB

A

www.VNMATH.com

www.vnmath.com

41

a) Ta co DMC = DME + CME = B + BDM , ma DME = B(gt) nen CME = BDM , ket hp vi B = C (ABC can tai A) suy ra BDM CME (g.g) 2BD BM = BD. CE = BM. CM = a

CM CE khong oi

b) BDM CME DM BD DM BD = = ME CM ME BM

(do BM = CM) DME DBM (c.g.c) MDE = BMD hay DM la tia phan giac cua BDE c) chng minh tng t ta co EM la tia phan giac cua DEC ke MH CE ,MI DE, MK DB th MH = MI = MK DKM = DIM DK =DI EIM = EHM EI = EH Chu vi AED la PAED = AD + DE + EA = AK +AH = 2AH (V AH = AK) ABC la tam giac eu nen suy ra CME cung la tam giac eu CH = MC

2 2a

AH = 1,5a PAED = 2 AH = 2. 1,5 a = 3a Bai 5: Cho tam giac ABC, trung tuyen AM. Qua iem D thuoc canh BC, ve ng thang song song vi AM, cat AB, AC tai E va F a) chng minh DE + DF khong oi khi D di ong tren BC b) Qua A ve ng thang song song vi BC, cat FE tai K. Chng minh rang K la trung iem cua FE Giai a) DE // AM DE BD BD = DE = .AM

AM BM BM (1)

DF // AM DF CD CD CD = DF = .AM = .AM AM CM CM BM

(2) T (1) va (2) suy ra

DE + DF = BD CD .AM + .AMBM BM

= BD CD BC+ .AM = .AM = 2AMBM BM BM

khong oi

b) AK // BC suy ra FKA AMC (g.g) FK KA = AM CM

(3)

EK KA EK KA EK KA EK KA EK KA = = = ED BD ED + EK BD + KA KD BD + DM AM BM AM CM

(2) (V CM = BM) T (1) va (2) suy ra FK EK

AM AM FK = EK hay K la trung iem cua FE

Bai 6: (e HSG huyen Thach ha nam 2003 2004) Cho hnh thoi ABCD canh a co 0A = 60 , mot ng thang bat ky qua C cat tia oi cua cac tia BA, DA tai M, N

KH

I

M

E

D

CB

A

K

F

E

D MCB

A

www.VNMATH.com

www.vnmath.com

42

a) Chng minh rang tch BM. DN co gia tr khong oi b) Goi K la giao iem cua BN va DM. Tnh so o cua goc BKD Giai a) BC // AN MB CM =

BA CN(1)

CD// AM CM AD = CN DN

(2)

T (1) va (2) suy ra 2MB AD = MB.DN = BA.AD = a.a = a

BA DN

b) MBD vaBDN co MBD = BDN = 1200 MB MB CM AD BD = =

BD BA CN DN DN (Do ABCD la hnh thoi

co 0A = 60 nen AB = BC = CD = DA) MBD BDN Suy ra 1 1M = B . MBD vaBKD co BDM = BDK va 1 1M = B nen 0BKD = MBD = 120 Bai 7: Cho hnh bnh hanh ABCD co ng cheo ln AC,tia Dx cat SC, AB, BC lan lt tai I, M, N. Ve CE vuong goc vi AB, CF vuong goc vi AD, BG vuong goc vi AC. Goi K la iem oi xng vi D qua I. Chng minh rang a) IM. IN = ID2 b) KM DM =

KN DN

c) AB. AE + AD. AF = AC2 Giai a) T AD // CM IM CI =

ID AI (1)

T CD // AN CI ID AI IN

(2)

T (1) va (2) suy ra IMID

= IDIN

hay ID2 = IM. IN

b) Ta co DM CM DM CM DM CM = = = MN MB MN + DM MB + CM DN CB

(3) T ID = IK va ID2 = IM. IN suy ra IK2 = IM. IN IK IN IK - IM IN - IK KM KN KM IM = = = =

IM IK IM IK IM IK KN IK KM IM CM CM =

KN ID AD CB (4)

T (3) va (4) suy ra KM DM = KN DN

c) Ta co AGB AEC AE AC= AB.AE = AC.AGAG AB

AB. AE = AG(AG + CG) (5) CGB AFC AF CG CG =

AC CB AD (v CB = AD)

AF . AD = AC. CG AF . AD = (AG + CG) .CG (6)

1

1 K

M

ND

CB

A

I

K

F

G

E

M

DC

BA N

www.VNMATH.com

www.vnmath.com

43

Cong (5) va (6) ve theo ve ta co: AB. AE + AF. AD = (AG + CG) .AG + (AG + CG) .CG AB. AE + AF. AD = AG2 +2.AG.CG + CG2 = (AG + CG)2 = AC2 Vay: AB. AE + AD. AF = AC2 Bai tap ve nha Bai 1 Cho Hnh bnh hanh ABCD, mot ng thang cat AB, AD, AC lan lt tai E, F, G Chng minh: AB AD AC + =

AE AF AG

HD: Ke DM // FE, BN // FE (M, N thuoc AC) Bai 2: Qua nh C cua hnh bnh hanh ABCD, ke ng thang cat BD, AB, AD E, G, F chng minh: a) DE2 = FE

EG. BE2

b) CE2 = FE. GE (Gi y: Xet cac tam giac DFE va BCE, DEC va BEG) Bai 3 Cho tam giac ABC vuong tai A, ng cao AH, trung tuyen BM, phan giac CD cat nhau tai mot iem. Chng minh rang a) BH CM AD. . 1

HC MA BD

b) BH = AC

www.VNMATH.com

www.vnmath.com

44

CHUYEN E 11 PHNG TRNH BAC CAO A.Muc tieu: * Cung co, on tap kien thc va ky nang giai cac Pt bac cao bang cach phan tch thanh nhan t * Khac sau ky nang phan tch a thc thanh nhan t va ky nang giai Pt B. Kien thc va bai tap: I. Phng phap: * Cach 1: e giai cac Pt bac cao, ta bien oi, rut gon e da Pt ve dang Pt co ve trai la mot a thc bac cao, ve phai bang 0, van dung cac phng phap phan tch a thc thanh nhan t e a Pt ve dang pt tch e giai * Cach 2: at an phu II. Cac v du: 1.V du 1: Giai Pt a) (x + 1)2(x + 2) + (x 1)2(x 2) = 12 ... 2x3 + 10x = 12 x3 + 5x 6 = 0 (x3 1) + (5x 5) (x 1)(x2 + x + 6) = 0

22

x = 1x - 1 = 0

x 11 23x + x + 6 = 0 x + 02 4

(V

21 23x + 02 4

vo nghiem)

b) x4 + x2 + 6x 8 = 0 (1) Ve phai cua Pt la mot a thc co tong cac he so bang 0, nen co mot nghiem x = 1 nen co nhan t la x 1, ta co (1) (x4 x3) + (x3 x2) + (2x2 2x) + (8x 8) = 0 ... (x 1)(x3 + x2 + 2x + 8) (x 1)[(x3 + 2x2) (x2 + 2x) + (4x 8) ] = 0 (x 1)[x2(x + 2) x(x + 2) + 4(x + 2) = 0 (x 1)(x + 2)(x2 x + 4) = 0 .... c) (x 1)3 + (2x + 3)3 = 27x3 + 8 x3 3x2 + 3x 1 + 8x3 + 36x2 + 54x + 27 27x3 8 = 0 - 18x3 + 33x2 + 57 x + 18 = 0 6x3 - 11x2 - 19x - 6 = 0 (2) Ta thay Pt co mot nghiem x = 3, nen ve trai co nhan t x 3: (2) (6x3 18x2) + (7x2 21x) + (2x 6) = 0 6x2(x 3) + 7x(x 3) + 2(x 3) = 0 (x 3)(6x2 + 7x + 2) = 0 (x 3)[(6x2 + 3x) + (4x + 2)] = 0 (x 3)[3x(2x + 1) + 2(2x + 1)] = 0 (x 3)(2x + 1)(3x + 2) ..... d) (x2 + 5x)2 2(x2 + 5x) = 24 [(x2 + 5x)2 2(x2 + 5x) + 1] 25 = 0 (x2 + 5x - 1)2 25 = 0 (x2 + 5x - 1 + 5)( (x2 + 5x - 1 5) = 0 (x2 + 5x + 4) (x2 + 5x 6) = 0 [(x2 + x) +(4x + 4)][(x2 x) + (6x 6)] = 0 (x + 1)(x + 4)(x 1)(x + 6) = 0 .... e) (x2 + x + 1)2 = 3(x4 + x2 + 1) (x2 + x + 1)2 - 3(x4 + x2 + 1) = 0

www.VNMATH.com

www.vnmath.com

45

(x2 + x + 1)2 3(x2 + x + 1)( x2 - x + 1) = 0 ( x2 + x + 1)[ x2 + x + 1 3(x2 - x + 1)] = 0 ( x2 + x + 1)( -2x2 + 4x - 2) = 0 (x2 + x + 1)(x2 2x + 1) = 0 ( x2 + x + 1)(x 1)2 = 0... f) x5 = x4 + x3 + x2 + x + 2 (x5 1) (x4 + x3 + x2 + x + 1) = 0 (x 1) (x4 + x3 + x2 + x + 1) (x4 + x3 + x2 + x + 1) = 0 (x 2) (x4 + x3 + x2 + x + 1) = 0 +) x 2 = 0 x = 2 +) x4 + x3 + x2 + x + 1 = 0 (x4 + x3) + (x + 1) + x2 = 0 (x + 1)(x3 + 1) + x2 = 0 (x + 1)2(x2 x + 1) + x2 = 0 (x + 1)2 [(x2 2.x. 1

2 + 1

4) + 3

4] + x2 = 0

(x + 1)2 21 3x + +

2 4

+ x2 = 0 Vo nghiem v (x + 1)2 21 3x + +

2 4 0 nhng

khong xay ra dau bang Bai 2: a) (x2 + x - 2)( x2 + x 3) = 12 (x2 + x 2)[( x2 + x 2) 1] 12 = 0 (x2 + x 2)2 (x2 + x 2) 12 = 0 at x2 + x 2 = y Th (x2 + x 2)2 (x2 + x 2) 12 = 0 y2 y 12 = 0 (y 4)(y + 3) = 0 * y 4 = 0 x2 + x 2 4 = 0 x2 + x 6 = 0 (x2 + 3x) (2x + 6) = 0 (x + 3)(x 2) = 0.... * y + 3 = 0 x2 + x 2 + 3 = 0 x2 + x + 1 = 0 (vo nghiem) b) (x 4)( x 5)( x 6)( x 7) = 1680 (x2 11x + 28)( x2 11x + 30) = 1680 at x2 11x + 29 = y , ta co: (x2 11x + 28)( x2 11x + 30) = 1680 (y + 1)(y 1) = 1680 y2 = 1681 y = 41 y = 41 x2 11x + 29 = 41 x2 11x 12 = 0 (x2 x) + (12x 12) = 0 (x 1)(x + 12) = 0..... * y = - 41 x2 11x + 29 = - 41 x2 11x + 70 = 0 (x2 2x. 11

2+121

4)+159

4 = 0

c) (x2 6x + 9)2 15(x2 6x + 10) = 1 (3) at x2 6x + 9 = (x 3)2 = y 0, ta co (3) y2 15(y + 1) 1 = 0 y2 15y 16 = 0 (y + 1)(y 15) = 0 Vi y + 1 = 0 y = -1 (loai) Vi y 15 = 0 y = 15 (x 3)2 = 16 x 3 = 4 + x 3 = 4 x = 7 + x 3 = - 4 x = - 1 d) (x2 + 1)2 + 3x(x2 + 1) + 2x2 = 0 (4) at x2 + 1 = y th (4) y2 + 3xy + 2x2 = 0 (y2 + xy) + (2xy + 2x2) = 0 (y + x)(y + 2x) = 0

www.VNMATH.com

www.vnmath.com

46

+) x + y = 0 x2 + x + 1 = 0 : Vo nghiem +) y + 2x = 0 x2 + 2x + 1 = 0 (x + 1)2 = 0 x = - 1 Bai 3: a) (2x + 1)(x + 1)2(2x + 3) = 18 (2x + 1)(2x + 2)2(2x + 3) = 72. (1) at 2x + 2 = y, ta co (1) (y 1)y2(y + 1) = 72 y2(y2 1) = 72 y4 y2 72 = 0 at y2 = z 0 Th y4 y2 72 = 0 z2 z 72 = 0 (z + 8)( z 9) = 0 * z + 8 = 0 z = - 8 (loai) * z 9 = 0 z = 9 y2 = 9 y = 3 x = ... b) (x + 1)4 + (x 3)4 = 82 (2) at y = x 1 x + 1 = y + 2; x 3 = y 2, ta co (2) (y + 2)4 + (y 2)4 = 82 y4 +8y3 + 24y2 + 32y + 16 + y4 - 8y3 + 24y2 - 32y + 16 = 82 2y4 + 48y2 + 32 82 = 0 y4 + 24y2 25 = 0 at y2 = z 0 y4 + 24y2 25 = 0 z2 + 24 z 25 = 0 (z 1)(z + 25) = 0 +) z 1 = 0 z = 1 y = 1 x = 0; x = 2 +) z + 25 = 0 z = - 25 (loai) Chu y: Khi giai Pt bac 4 dang (x + a)4 + (x + b)4 = c ta thng at an phu y = x + a + b

2c) (4 x)5 + (x 2)5 = 32 (x 2)5 (x 4)5 = 32 at y = x 3 x 2 = y + 1; x 4 = y 1; ta co: (x 2)5 (x 4)5 = 32 (y + 1)5 - (y 1)5 = 32 y5 + 5y4 + 10y3 + 10y2 + 5y + 1 (y5 - 5y4 + 10y3 - 10y2 + 5y - 1) 32 = 0 10y4 + 20y2 30 = 0 y4 + 2y2 3 = 0 at y2 = z 0 y4 + 2y2 3 = 0 z2 + 2z 3 = 0 (z 1)(z + 3) = 0 ........ d) (x - 7)4 + (x 8)4 = (15 2x)4 at x 7 = a; x 8 = b ; 15 2x = c th - c = 2x 15 a + b = - c , Nen (x - 7)4 + (x 8)4 = (15 2x)4 a4 + b4 = c4 a4 + b4 - c4 = 0 a4 + b4 (a + b)4 = 0

4ab(a2 + 32

ab + b2) = 0 2

23 74ab a + b + b 4 16

= 0 4ab = 0

(V 2

23 7a + b + b4 16

0 nhng khong xay ra dau bang) ab = 0 x = 7; x = 8

e) 6x4 + 7x3 36x2 7x + 6 = 0 2 21 16 x 7 x - 36 0x x

(V x = 0 khong la nghiem). at 1x - x

= y 2 21x x = y2 + 2 , th

www.VNMATH.com

www.vnmath.com

47

22

1 16 x 7 x - 36 0x x

6(y2 + 2) + 7y 36 = 0 6y2 + 7y 24 = 0

(6y2 9y) + (16y 24) = 0 (3y + 8 )(2y 3) = 0

+) 3y + 8 = 0 y = - 83 1x -

x = - 8

3 ... (x + 3)(3x 1) = 0

x = - 3x + 3 = 013x - 1 = 0 x = 3

+) 2y 3 = 0 y = 32 1x -

x = 3

2 ... (2x + 1)(x 2) = 0

x = 2x - 2 = 012x + 1 = 0 x = - 2

Bai 4: Chng minh rang: cac Pt sau vo nghiem a) x4 3x2 + 6x + 13 = 0 ( x4 4x2 + 4) +(x2 + 6x + 9) = 0 (x2 2)2 + (x + 3)2 = 0 Ve trai (x2 2)2 + (x + 3)2 0 nhng khong ong thi xay ra x2 = 2 va x = -3 b) x6 + x5 + x4 + x3 + x2 + x + 1 = 0 (x 1)( x6 + x5 + x4 + x3 + x2 + x + 1) = 0 x7 1 = 0 x = 1 x = 1 khong la nghiem cua Pt x6 + x5 + x4 + x3 + x2 + x + 1 = 0 Bai tap ve nha: Bai 1: Giai cac Pt a)(x2 + 1)2 = 4(2x 1) HD: Chuyen ve, trien khai (x2 + 1)2, phan tch thanh nhan t: (x 1)2(x2 + 2x + 5) = 0 b) x(x + 1)(x + 2)(x + 3) = 24 (Nhan 2 nhan t vi nhau, ap dung PP at an phu) c) (12x + 7)2(3x + 2)(2x + 1) = 3 (Nhan 2 ve vi 24, at 12x + 7 = y) d) (x2 9)2 = 12x + 1 (Them, bt 36x2) e) (x 1)4 + (x 2)4 = 1 ( at y = x 1,5; s: x = 1; x = 2) f) (x 1)5 + (x + 3)5 = 242(x + 1) (at x + 1 = y; s:0; -1; -2 ) g) (x + 1)3 + (x - 2)3 = (2x 1)3 at x + 1 = a; x 2 = b; 1 - 2x = c th a + b + c = 0 a3 + b3 + c3 = 3abc h) 6x4 + 5x3 38x2 + 5x + 6 = 0 (Chia 2 ve cho x2; at y = 1x +

x )

i) x5 + 2x4 + 3x3 + 3x2 + 2x + 1 = 0 (Ve trai la a thc co tong cac he so bac chan bang tong cac he so bac le...) Bai 2: Chng minh cac pt sau vo nghiem a) 2x4 10x2 + 17 = 0 (Phan tch ve trai thanh tong cua hai bnh phng) b) x4 2x3 + 4x2 3x + 2 = 0 (Phan tch ve trai thanh tch cua 2 a thc co gia tr khong am....)

www.VNMATH.com

www.vnmath.com

48

CHUYEN E 12 VE NG THANG SONG SONG E TAO THANH CAC CAP OAN THANG TY LE

A. Phng phap: Trong cac bai tap van dung nh l Talet. Nhieu khi ta can ve them ng phla mot ng thang song song vi mot ng thang cho trc,. ay la mot cach ve ng phu hay dung, v nh o ma tao thanh c cac cap oan thang t le B. Cac v du: 1) V du 1: Tren cac canh BC, CA, AB cua tam giac ABC, lay tng ng cac iem P, Q, R sao cho ba ng thang AP, BQ, CR cat nhau tai mot iem. Chng minh: AR BP CQ. . 1

RB PC QA (nh l Ce va)

Giai Qua A ke ng thang song song vi BC cat cac ng thang CR, BQ tai E, F. Goi O la giao iem cua AP, BQ, CR ARE BRC AR AE =

RB BC (a)

BOP FOA BP OP = FA OA

(1)

POC AOE PC PO = AE AO

(2)

T (1) va (2) suy ra: BP PC BP FA = FA AE PC AE

(b)

AQF CQB CQ BC = AQ FA

(c)

Nhan (a), (b), (c) ve theo ve ta co: AR BP CQ AE FA BC. . . . 1RB PC QA BC AE FA

* ao lai: Neu AR BP CQ. . 1RB PC QA

th bai ng thang AP, BQ, CR ong quy 2) V du 2: Mot ng thang bat ky cat cac canh( phan keo dai cua cac canh) cua tam giac ABC tai P, Q, R. Chng minh rang: RB.QA.PC 1

RA.CQ.BP (nh l Me-ne-la-uyt)

Giai: Qua A ke ng thang song song vi BC cat PR tai E. Ta co RAE RBP RB BP =

RA AE (a)

O

FE

R Q

CPB

A

E

R

Q

CPB

A

www.VNMATH.com

www.vnmath.com

49

AQE CQP QA AE = QC CP

(b)

Nhan ve theo ve cac ang thc (a) va (b) ta co RB QA BP AE. = .RA QC AE CP

(1)

Nhan hai ve ang thc (1) vi PCBP

ta co: RB PC QA BP AE PC. . = . . 1RA BP QC AE CP BP

ao lai: Neu RB.QA.PC 1RA.CQ.BP

th ba iem P, Q, R thang hang 3) V du 3: Cho tam giac ABC, trung tuyen AM. Goi I la iem bat ky tren canh BC. ng thang qua I song song vi AC cat AB K; ng thang qua I song song vi AB cat AC, AM theo th t D, E. Chng minh DE = BK Giai Qua M ke MN // IE (N AC).Ta co: DE AE DE MN = MN AN AE AN

(1) MN // IE, ma MB = MC AN = CN (2) T (1) va (2) suy ra DE MN

AE CN (3)

Ta lai co MN CN MN ABAB AC CN AC

(4)

T (4) va (5) suy ra DE ABAE AC

(a)

Tng t ta co: BK ABKI AC

(6) V KI // AC, IE // AC nen t giac AKIE la hnh bnh hanh nen KI = AE (7) T (6) va (7) suy ra BK BK AB

KI AE AC (b)

T (a) va (b) suy ra DE BKAE AE

DE = BK 4) V du 4: ng thang qua trung iem cua canh oi AB, CD cua t giac ABCD cat cac ng thang AD, BC theo th t I, K. Chng minh: IA . KC = ID. KB Giai Goi M, N theo th t la trung iem cua AB, CD Ta co AM = BM; DN = CN Ve AE, BF lan lt song song vi CD AME = BMF (g.c.g) AE = BF

N

D

I M

E

K

CB

A

F

E

I

K

M

ND C

B

A

www.VNMATH.com

www.vnmath.com

50

Theo nh l Talet ta co: IA AE BF = ID DN CN

(1)

Cung theo nh l Talet ta co: KB BF = KC CN

(2)

T (1) va (2) suy ra IA KB =ID KC

IA . KC = ID. KB 5) V du 5: Cho xOy , cac iem A, B theo th t chuyen ong tren cac tia Ox, Oy sao cho

1 1 1 + OA OB k

(k la hang so). Chng minh rang AB luon i qua mot iem co nh Giai Ve tia phan giac Oz cua xOy cat AB C. ve CD // OA (D OB) DOC = DCO = AOC COD can tai D DO = DC Theo nh l Talet ta co CD BD CD OB - CD =

OA OB OA OB

CD CD 1 1 11OA OB OA OB CD

(1)

Theo gia thiet th 1 1 1 + OA OB k

(2) T (1) va (2) suy ra CD = k , khong oi Vay AB luon i qua mot iem co nh la C sao cho CD = k va CD // Ox , D OB 6) V du 6: Cho iem M di ong tren ay nho AB cua hnh thang ABCD, Goi O la giao iem cua hai canh ben DA, CB. Goi G la giao iem cua OA va CM, H la giao iem cua OB va DM. Chng minh rang: Khi M di ong tren AB th tong OG OH +

GD HC

khong oi Giai Qua O ke ng thang song vi AB cat CM, DM theo th t I va K. Theo nh l Talet ta co: OG OIGD CD

; OH OKHC CD

OG OH OI OK IK + GD HC CD CD CD

OG OH IK + GD HC CD

(1) Qua M ve ng thang vuong goc vi AB cat IK, CD theo th t P va Q, ta co: IK MP FOCD MQ MQ

khong oi v FO la khoang cach t O en AB, MQ la ng cao cua hnh thang nen khong oi (2) T (1) va (2) suy ra OG OH FO +

GD HC MQ khong oi

Q

P

F

KI

HG

M

O

D C

BA

z

O

y

x

DC

B

A

www.VNMATH.com

www.vnmath.com

51

www.vnmath.com

7) V du 7: Cho tam giac ABC (AB < AC), phan giac AD. Tren AB lay iem M, tren AC lay iem N sao cho BM = CN, goi giao iem cua CM va BN la O, T O ve ng thang song song vi AD cat AC, AB tai E va F. Chng minh rang: AB = CF; BE = CA Giai. AD la phan giac nen BAD = DAF EI // AD BAD = AEF (goc ong v) Ma DAF OFC (ong v); AFE = OFC (oi nh) Suy ra AEF AFE AFE can tai A AE =AF (a) Ap dung nh l Talet vao ACD , vi I la giao iem cua EF vi BC ta co CF CI CF CA =

CA CD CI CD (1)

AD la phan giac cua BAC nen CA BACD BD

(2)

T (1) va (2) suy ra CF BACI BD

(3) Ke ng cao AG cua AFE . BP // AG (P AD); CQ // AG (Q OI) th BPD = CQI = 900 Goi trung iem cua BC la K, ta co BPK = CQK (g.c.g) CQ = BP BPD = CQI (g.c.g) CI = BD (4) Thay (4) vao (3) ta co CF BA

BD BD CF = BA (b)

T (a) va (b) suy ra BE = CA Bai tap ve nha 1) Cho tam giac ABC. iem D chia trong BC theo t so 1 : 2, iem O chia trong AD theo t so 3 : 2. goi K la giao iem cua BO va AC. Chng minh rang KA

KC khong oi

2) Cho tam giac ABC (AB > AC). Lay cac iem D, E tuy y th t thuoc cac canh AB, AC sao cho BD = CE. Goi giao iem cua DE, BC la K, chng minh rang : T so KE

KD khong oi khi D, E thay oi tren AB, AC

(HD: Ve DG // EC (G BC).

G

P O

KI

N

D

Q

CB

M

A

F

E

www.VNMATH.com

www.vnmath.com

52

CHUYEN E 13 BO E HNH THANG VA CHUM NG THANG ONG QUY

A. Kien thc: 1) Bo e hnh thang: Trong hnh thang co hai ay khong bang nhau, ng thang i qua giao iem cua cac ng cheo va i qua giao iem cua cac ng thang cha hai canh ben th i qua trung iem cua hai ay Chng minh: Goi giao iem cua AB, CD la H, cua AC, BD la G, trung iem cua AD, BC la E va F Noi EG, FG, ta co: ADG CBG (g.g) , nen : AD AG 2AE AG AE AGCB CG 2CF CG CF CG

(1) Ta lai co : EAG FCG (SL trong ) (2) T (1) va (2) suy ra : AEG CFG (c.g.c) Do o: AGE CGF E , G , H thang hang (3) Tng t, ta co: AEH BFH AHE BHF H , E , F thang hang (4) T (3) va (4) suy ra : H , E , G , F thang hang 2) Chum ng thang ong quy: Neu cac ng thang ong quy cat hai ng thang song song th chung nh ra tren hai ng thang song song ay cac oan thang tng ng t le Neu m // n, ba ng thang a, b, c ong quy O chung cat m tai A, B, C va cat n tai A, B, C th AB BC AC = A'B' B'C' A'C'

hoac AB A'B' AB A'B' = ; BC B'C' AC A'C'

* ao lai: + Neu ba ng thang trong o co hai ng thang cat nhau, nh ra tren hai ng thang song song cac cap oan thang tng ng t le th ba ng thang o ong quy + Neu hai ng thang b cat bi ba ng thang ong quy tao thanh cac cap oan thang tng ng t le th chung song song vi nhau B. Ap dung: 1) Bai 1: Cho t giac ABCD co M la trung iem CD, N la trung iem CB. Biet AM, AN cat BD thanh ba oan bang nhau. Chng minh rang ABCD la hnh bnh hanh Giai

////

//

H

G

E

F

D

CB

A

cba

O

n

m

A' B' C'

CBA

www.VNMATH.com

www.vnmath.com

53

Goi E, F la giao iem cua AM, AN vi BD; G, H la giao iem cua MN vi AD, BD MN // BC (MN la ng trung bnh cua BCD) T giac HBFM la hnh thang co hai canh ben ong quy tai A, N la trung iem cua ay BF nen theo bo e hnh thang th N la trung iem cua ay MH MN = NH (1) Tng t : trong hnh thang CDEN th M la trung iem cua GN GM = MN (2) T (1) va (2) suy ra GM = MN = NH Ta co BNH = CNM (c.g.c) BHN = CMN BH // CM hay AB // CD (a) Tng t: GDM = NCM (c.g.c) DGM = CNM GD // CN hay AD // CB (b) T (a) va (b) suy ra t giac ABCD co cac cap canh oi song song nen la hnh bnh hanh 2) Bai 2: Cho ABC co ba goc nhon, trc tam H, mot ng thang qua H cat AB, AC th t ta P, Q sao cho HP = HQ. Goi M la trung iem cua BC. Chng minh: HM PQ Giai Goi giao iem cua AH va BC la I T C ke CN // PQ (N AB), ta chng minh MH CN HM PQ T giac CNPQ la hnh thang, co H la trung iem PQ, hai canh ben NP va CQ ong quy tai A nen K la trung iem CN MK la ng trung bnh cua BCN MK // CN MK // AB (1) H la trc tam cua ABC nen CHA B (2) T (1) va (2) suy ra MK CH MK la ng cao cuaCHK (3) T AH BC MCHK MI la ng cao cua CHK (4) T (3) va (4) suy ra M la trc tam cua CHK MHCN MHPQ 3) bai 3: Cho hnh ch nhat ABCD co M, N th t la trung iem cua AD, BC. Goi E la mot iem bat ky thuoc tia oi cua tia DC, K la giao iem cua EM va AC. Chng minh rang: NM la tia phan giac cua KNE Giai Goi H la giao iem cua KN va DC, giao iem cua AC va MN la I th IM = IN Ta co: MN // CD (MN la ng trung bnh cua hnh

H

G

F

E

N

M

D

CB

A

I

K

N

M

Q

P

H

CB

A

////I

H E

N M

K

DC

B A

www.VNMATH.com

www.vnmath.com

54

ch nhat ABCD) T giac EMNH la hnh thang co hai canh ben EM va HN ong quy tai K va I la trung iem cua MN nen C la trung iem cua EH Trong ENH th NC va la ng cao, va la ng trung tuyen nen ENH can tai N NC la tia phan giac cua ENH ma NC MN (Do NM BC MN // AB) NM la tia phan giac goc ngoai tai N cua ENH Vay NM la tia phan giac cua KNE Bai 4: Tren canh BC = 6 cm cua hnh vuong ABCD lay iem E sao cho BE = 2 cm. Tren tia oi cua tia CD lay iem F sao cho CF = 3 cm. Goi M la giao iem cua AE va BF. Tnh AMC Giai Goi giao iem cua CM va AB la H, cua AM va DF la G Ta co: BH AB BH 6 =

CF FG 3 FG

Ta lai co AB BE 2 1 = = CG = 2AB = 12