Cac Phuong Phap Giai Bat Phuong Trinh(2)

8/8/2019 Cac Phuong Phap Giai Bat Phuong Trinh(2)

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CHNG 2: BT PHNG TRNH1.Phng php s dng tnh n iu ca hm s:Th d 128: Gii bt phng trnh: (1)542x9x >+++

Li gii:

t f(x) = VT(1), c f(x) xc nh: (1) (2x2x

9x

042 x

09x

++

f(x) xc nh, lin tc trn (*) c: 042x2

1

9x2

1(x)f' >

++

+= vi x >

-2

nn f(x) ng bin trn (*). Do : 0x2x

0xf ( 0 )f ( x )( 1 ) >

>>

Vy bt phng trnh c nghim: 0x > .

Th d 129: Gii bt phng trnh: (1)55xx +Li gii:

t f(x) = VT(1), c f(x) xc nh: (1) (5x5x

0x

05x

0x

f(x) xc nh, lin tc trn (*) c: 0

5x2

1

x2

1(x)f

' >

+= vi 5x >


2/40

nn f(x) ng bin trn (*).Do : 5x5x

5xf ( 5 )f ( x )( 1 ) =

Vy bt phng trnh c nghim: 5x = .

Th d 130: Gii bt phng trnh: (1)38532 xxx ++Li gii:

t f(x) = VT(1), c f(x) xc nh v lin tc vi mi x R c:0ln55ln33ln22(x)f xxx' >++= vi mi x R nn f(x) ng bin trn

(*).Do

2xRx

2x

f ( 2 )f ( x )( 1 )

Vy bt phng trnh c nghim: 2x .

Th d 131: (NTA-2000) Gii bt phng trnh:(1)22)(4log1)(2log x3

x

2 +++Li gii:

t f(x) = VT(1),c f(x) xc nh,lin tc vi mi Rx c:

02)ln3(4

ln44

1)ln2(2

ln22(x)f

x

x

x

x' >

++

+= vi mi Rx

nn f(x) ng bin trn (*).Do : 0xRx

0xf ( 0 )f ( x )( 1 )

Vy bt phng trnh c nghim: 0x .

Th d 132: (TL-2000) Gii bt phng trnh: (1)2x5x-32x


3/40

Ta c f(x) xc nh khi v ch khi (

2

5x2

02 x5

0x-3

02x

+

f(x) xc nh, lin tc trn (*) c: 02x-52

1

x-32

1

2x2

1(x)f' >++

+=

vi2

5x2


4/40

t f(x) = VT(2), c f(x) xc nh khi v ch khi:

++

+++

0x4

08x-2 ) (( x

0x4

01 66 x3 x2 x 223

>++

4x

08x-2 x( d o02x 2 (*)4x2

f(x) xc nh, lin tc trn (*) c:

0x42

1

166x3x2x2

66x-6x

(x)f 23

2'

>++++

+

= vi 4x2


5/40

Th d 136: Gii h thc: log x (x + 1) = lg1,5 (1)Li gii:

iu kin: 0 < x 1- Xt 0 < x < 1 khi logx(x+1) < logx1 = 0 < lg1,5. Vy phng trnh (1) khngc nghim trong khong ny

- Xt 1 < x < + khi logx(x+1) > logxx = 1 > lg1,5. Vy phng trnh (1)khng c nghim trong khong nyTm li (1) v nghim.

Th d 137: Gii h thc 2x

24xx3 2

>

>

Kt hp vi iu kin (*) ta c3

4x

7

9< .


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Th d 138: Gii h thc

>+


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- Vi 2x > th x2 4 > 0 v x 2 > 0. Do 133 04x2

=> (v hm ngbin)nn VT(1) > 1 = VP(1). Bt phng trnh khng c nghim trong khong trn- Vi 2x < th x2 4 < 0 v x 2 < 0. Do 133 04x

2

= 0 nn ( ) 02x;11x 1x353 -1- Nu x > 1 th ( )2x2log)xx(log

)1x(2

)1x(xlogVP 2

2

22 =

= m

VT = 22x-2- xx22 . Do : (1) xx2

2 + ( )2x2log)xx(log 222 = +22x-2(1/)Xt hm s f(x) = 2t + log2t xc nh lin tc trn R+ v:

f/(x) = t.ln2 + 2ln.t

1

< 0 nn f(x) nghch bin trn R+

(1/) x2 x = 2x 2 x2 3x + 2 = 0 x = 1 (loi); x = 2 (tha mn).Vy phng trnh c nghim x = 2.

Th d 143: Gii phng trnh 182x6xx 2 =+++ (1)Li gii:

iu kin: x + 2 0 x 2. t f(x) = 2x6xx 2 +++ c f(x) xc nh,

lin tc trn [ )+ ;2 v f/(x) = 2x + 1 +2x

3

+- Nu x 0 th f/(x) > 0 nn VT(1) l hm ng bin m VP(1) = const do phng trnh c nghim duy nht x = 2

- Nu 2 x < 0 th VT(1) < 18 = VP(1) nn phng trnh khng c nghimtrong khong trn .Tm li phng trnh c nghim duy nht x = 2.

Th d 144: Gii phng trnh: x4 + x3 + 5 2x + = 2 + 5 2 (1)


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Li gii:t f(x) = 1x5xx 34 +++ c f(x) xc nh lin tc trn [ )+ ;1

f/(x) =1x2

5x3x4 23

+++

- Nu x 0 th f/(x) > 0 nn f(x) ng bin do VT(1) ng bin m

VP(1) = const. V vy x = 1 l nghim duy nht ca phng trnh- Nu 1 x < 0 ta thy VT(1) < 6 < VP(1).Vy phng trnh c nghim duy nht x = 1.

3: Phng php hm lin tc:

Th d 145: Gii bt phng trnh )1(0xx4

3x24

xtg

2

+=

+=

+=

=

0n;k

1xnx21

x31n1

1xkx21

x31k1

( )( )

( )

+=

+=

++=

>=

x31k1

1xkx21

x31x31k1

0nk

=

=

+=

+=

=

3

1x

5

1

x

x312

1xx422

1k

(loi)

Vy h cho v nghim.

Th d 175: Gii phnh trnh ( )11x2x3x1x 2 +=++Li gii:

iu kin: (3x10x30x1

+

. t ( ) ( )1;xv;x3;x1u +

p dng BT thc ( )n21n

1i

i

n

1i

i a..........aa:aa

===

c:

( ) ( )1VP1x2vuv.ux3x1x1VT2

=+===++=


30/40

iu c ngha l:

=

=+ =

kx3

kx1vuv.uv.u

Ta c:

( )

+=

=

=+

21x

1x

xx3x1

0x)1(

2

Vy phng trnh c nghim 21x;1x +== .

Th d 176: Gii phng trnh: 4x4xx 23 ++++ = x x + 2Li gii:

iu kin x 0 v x 3 + x 2 + 4x + 4 0 x 0 (*)

tu (x; 2) v

v ( x ; 1). Ta c:

u .

v |

u |.|

v |

hay x x + 2 4x 2 + . 1x + 4x4xx 23 ++++

ng thc xy ra khiu //

v x

x= 2 x = 4 (tho mn iu kin (*))

Vy nghim ca phng trnh l x = 4.

Th d 177: Gii phng trnh:

=++=++

7z2yx

1zyx222

444

Li gii:

tu (x

2 ; y 2 ; z 2 );v (1; 1; 2), c

u .

v |

u | |

v |

hay x 2 + y 2 +2z 2 444 zyx ++ 7 6 (iu ny l v l)Vy phng trnh v nghim.

Th d 179: Gii phng trnh ( ) 1x52x2 32 +=+ (1)

Li gii:t 1xv;1xxu 2 +=+= th u.v = x3 + 1 v u + v = x2 + 2 nn iu kin (1) xc nh l v 0 (V u lun ln hn 0). Vi iu kin :


31/40

(1) tr thnh 2(u + v) = 5 uv u

v5

u

v22 =+ (u 0)

( )

=

=

2

1

u

v

)(2u

v

* V ( ) ta c:

(


32/40

2sinx(sinx cox)(sinx + cosx) + (sinx cosx) = 0 (sinx cosx)(2sin2x + 2sinxcosx + 1) = 0

=++

=

)(01xcosxsin2xsin2

)(0xcosxsin

2

( ) )Zk(k4

x

k2x2

x

k2x2xx

2s i nxc o sxs i n +

=

++

=

+

=

==

( ) ( )

=

=

=

=+=++

0xs i

0xc o

0xs i n2

0xc o sxs i n0xs i n2xc o sxs i n

2

22

(v nghim)

Vy phng trnh c nghim l ( )Zkk4

x +

= .

Th d 181: Gii phng trnh sau

( ) ( ) ( )323232 1x2x33x5x24x3x =+++Li gii:

t b3x5x2;a4x3x22

=+=+ th a + b = 3x2

2x 1 v (1) tr thnha3 + b3 = (a + b)3

3ab(a + b) = 0

=

=+

=+

01x2x3

03x5x2

04x3x

2

2

2

2

3x1x

3

1x4x ====

Vy phng trnh c tp nghim l:

=

2

3;1;

3

1;4S .

Th d 182: Gii h:

+=+ =+2255

33

yxyx 1yx(I)

Li gii:


33/40

(I)

( ) ( )

++=+

=+

332255

33

yxyxyx

)1(1yx

( )

=+

=+

0yxyx

1yx

22

33

=+

==

=+

0yx

0y

0x

1yx 33

+) Vi x = 0 thay vo (1) ta c y = 1+) Vi y = 0 thay vo (1) ta c x = 1+) Vi x + y = 0 x = y thay vo (1) ta c y3 + y3 = 1 0 = 1 (v l)Vy h c cc cp nghim (x; y) l (0; 1); (1; 0).

Th d 183: Gii h

=

=

2)yx(x y

7yx 33

Li gii:

H bi cho tng ng vi

( )

=

=

)2(0)yx(x y7yx2

)1(7yx

33

33

(2) (x y)(2x2 + 2y2 5xy) = 0

=+

=

0x y5y2x2

0yx

22

- Nu x y = 0 x = y thay vo (1) ta c 32

7yx ==

- Nu 2x2 + 2y2 - 5xy = 0 (3)+) Vi y = 0 ta c x = 0 thay vo (1) thy v l

+) Vi y 0 chia hai v ca (3) cho y2 ta c:

=

==+

2

1

y

x

2y

x

02y

x5

y

x2

2

Khi x = 2y th (1) 8y3 y3 = 7 7y3 = 7 y = 1 nn x = 2Khi y = 2x tng t ta c x = 1; y = 2

Vy h c cc cp nghim (x; y) l )2,1();1;2();2

7;

2

7( 33 .


34/40


35/40

Li gii:iu kin: x, y 0. Vi iu kin

(I)

=

=

x4x y3y

y4x y3x2

2

=

=

y4x y3x

)xy(4yx2

22

=

=++

y4x y3x

0)4yx) (yx(2

==++

==

y4x y3x

04yx

y4x y3x

0yx

2

2

==

====

2yx0yx2yx

(Loi do x, y 0)

Vy h c nghim duy nht x = y = 2.

Th d 186: (VMO-96) Gii h

=+

=+

+

24)yx

1

1(y7

2)yx

11(x3

(I)

Li gii:D thy nu (x; y) l nghim ca h (I) th x; y > 0


36/40

Do (I)

=+

+=

=+

=+

+

(y7

22

x3

1

yx

1

(

y7

22

3 x

11

y7

24

yx

11

x3

2

yx

11

Nhn tng v ca (1) v (2) ta c:

y7

8

x3

1

yx

1=

+ 21xy = (x + y)(7y 24x) 7y2 38xy 24x2 = 0

(y 6x)(7y + 4x) = 0 y = 6x (Do x; y > 0)

Thay y = 6x vo (1) ta cx42

22

x3

11 +=

189

211257x

+=

63

2124114y

+= (th li thy tho mn)

Vy h c nghim (x; y) = (63

2124114;

189

211257 ++ )

Th d 187: Gii bin lun theo a phng trnh ax = (a 1)x + 1 (a > 0)Li gii:

(1) 01x)1a(a)x(fx

== . Nhn xt rng:ao (a 1).0 1 = 1 1 = 0; a1 (a - 1).1 1 = 0 nn (1) c t nht hai nghimphn bit x = 0; x = 1.Gi s (1) c nhiu hn hai nghim th khi do f(x) xc nh v lin tc trn Rnn theo h qu nh l rolle ta c: f/(x) c t nht 2 nghim.Tng t f//(x) c t nht 1 nghim m f/(x) = axlna (a 1)f//(x) = ax(lna)2 > 0 Rx (iu ny tri vi kt qu trn)Vy 0a > phng trnh (1) c 2 nghim phn bit x = 0; x = 1.

Th d 188: Gii phng trnh: 3x + 2x = 3x 2Li gii:

Ta c: (2) f(x) = 3x + 2x 3x 2 = 0. Nhn xt rng:

3o + 2o 3.0 2 = 1 + 1 2; 31 + 21 3.1 2 = 5 5 = 0 nn (2) c t nht hainghim phn bit x = 0; x = 1.Gi s (2) c nhiu hn hai nghim th khi do f(x) xc nh; lin tc trn Rnn theo h qu nh l rolle ta c: f/(x) c t nht hai nghim.Tng t f//(x) c t nht 1 nghim m f/(x) = 3xln3 + 2xln2 3f//(x) = 3x(ln3)2 + 2x(ln2)2 > 0 Rx (iu ny tri vi kt qu trn)


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Vy phng trnh cho c hai nghim x = 0; x = 1.

Th d 189: Chng minh rng vi nN*; n chn v vi p; qR th phngtrnh xn + px + q = 0 (3) khng th c qu hai nghim thc phn bit

Li gii:Gi s (3) c nhiu hn hai nghim thc phn bit. Khi do f(x) = xn + px + qxc nh lin tc trn R nn theo h qu nh l Rolle ta c phng trnh f/(x) = 0c t nht hai nghim.Tng t f//(x) = 0 c t nht 1 nghim m f/(x) = n.xn 1 + p; f//(x) = n(n 1)xn 2

Do n N*; n chn nn n(n 1) > 0 v n 2 chnf//(x) > 0 xR (iu ny tri vi kt qu trn)Vy (3) khng th c qu hai nghim thc phn bit (pcm).

Th d 190: Chng minh rng nu a, b, c, d i mt khc nhau th phng trnh(x a)(x b)(x c) + (x d)(x b)(x c) + (x a)(x d)(x c) + (x a)(x b) .(x d) = 0 (4) lun c ba nghim phn bit

Li gii:t f(x) = VT(4) v F(x) = (x a)(x b)(x c)(x d) xc nh lin tc trn R, doF(a) = F(b) = F(c) = F(d) = 0 nn phng trnh F(x) = 0 c 4 nghim phn bit.Theo h qu nh l Rolle c F/(x) = 0 c t nht 3 nghim phn bit mF/(x) = f(x) f(x) c t nht 3 nghim phn bit.Nhng f(x) l a thc bc 3 nn n c nhiu nht 3 nghim phn bitVy (4) c ng 3 nghim phn bit (pcm).

Th d 191: Chng minh rng vi a, b, cR phng trnhacos3x + bcos2x + ccosx + sinx = 0 (5) lun c nghim x [ ]2;0 (*)

Li gii:

t ( ) 1cosx2sin2

bx3sin

3

axF +++= ta c F(x) xc nh v lin tc trn

[ ]2;0 ; kh vi ( )2;0 m F(0) = F(2) = 0 a;b;c R nn theo nh l Rollephng trnh F/(x) = 0 lun c nghim x (0;2) [ ] 2;0 m F/(x) = VT(5)Vy (5) lun c nghim x [ ]2;0 .

Th d 192: Chng minh rng vi a, b, c, d, e R phng trnh sau lun cnghim F(x) = acos6 + bcos5 + csin4x + dcos3x + esinx = 0

Li gii:

t ( ) e4

c

6

axcosex3sin

3

dx4cos

4

cx5sin

5

bx6sin

6

axF +++++=

Ta c F(x) xc nh v lin tc trn [ ]2;0 ; kh vi ( )2;0 m F(0) = F(2) =0 a, b, c, d, e R nn theo nh l Rolle phng trnh F/(x) = 0 lun c t nht

mt nghim x (0;2) [ ] 2;0 M F/(x) = f(x)Vy phng trnh cho lun c nghim.

Th d 193: Cho f(x) = anxn + an-1xn-1 + . + a1x + ao. Chng minh rng nu


38/40

nN* sao cho 0m

a

1m

a....

1nm

a

nm

a o111nn =++

+++

++

th phng trnh

f(x) = 0 lun c nghim x (0;1).Li gii:

t 0xm

ax

1m

a....x

1nm

ax

nm

a)x(F mo1m11nm1nnmn

=+++++++=

+++

Ta c F(x) xc nh v lin tc trn [ ]1;0 ; kh vi ( )1;0 v c F(0) = F(1) = 0nn theo nh l Rolle phng trnh F/(x) = 0 lun c t nht mt nghim x(0;1)M F/(x) = an x n + m 1 + an - 1 x n + m 2 + + a1 x m + aox m 1

= x m 1 (an x n + a n 1 x n - 1 +.+a1x + ao)Nu gi xo (0; 1) l mt nnghim ca phng trnh F/(x) = 0 th:

xm-1(anxn + an-1xn-1 +.+ a1x + ao) = 0 f(xo) = 0 (xo(0;1))Vy phng trnh f(x) = 0 lun c nghim (x (0; 1)).

Th d 194: Gii bt phng trnh0

1x32

x5x5lg

)x(fx

0 (iu ny tri vi kt qu trn)

Vy phng trnh cho c khng qu 3 nghim (pcm).

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Cac Phuong Phap Giai Bat Phuong Trinh(2)