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5/24/2018 Bt Hsg Tinh 2010
1/46
TI LIU BI DNG HC SINH GII
Phm Kim Chung THPT NG THC HA T : 0984.333.030 Mail :[email protected] Tr.
1
1
S GD&T NGH AN TRNG THPT NG THC HA
MT S BITON CHN LC BI DNGHC SINH GIIMN TON
VIT BI: PHM KIM CHUNG THNG 12 NM 2010
PHN MC LC Tr
I PHNG TRNH BPT HPT CC BI TON LIN QUAN N O HM
II PHNG TRNH HM VA THC
III BT NG THC V CC TR
IV GII HN CA DY S
V HNH HC KHNG GIAN
VI T LUYN V LI GII
DANH MC CC TI LIU THAM KHO
1. Cc din n :www.dangthuchua.com,www.math.vn,www.mathscope.org,www.maths.vn,www.laisac.page.www.diendantoanhoc.net,www.k2pi.violet.vn ,www.nguyentatthu.violet.vn,
2. thi HSG Quc Gia, thi HSG cc Tnh Thnh Ph trong nc, thi Olympic 30 -43. B sch : Mt s chuyn bi dng hc sinh gii ( Nguyn Vn Mu Nguyn Vn Tin ) 4. Tp ch Ton Hc v Tui Tr 5. B sch : CC PHNG PH P GII ( Trn Phng - L Hng c )6. B sch : 10.000 BI TON S CP (Phan Huy Khi )7. B sch : Ton nng cao ( Phan Huy Khi )8. Gii TON HNH HC 11 ( Trn Thnh Minh )9. Sng to Bt ng thc ( Phm Kim Hng )10. Bt ng thc Suy lun v khm ph ( Phm Vn Thun ) 11. Nhng vin kim cng trong Bt ng thc Ton hc ( Trn Phng ) 12. 340 bi ton hnh hc khng gian ( I.F . Sharygin ) 13. Tuyn tp 200 Bi thi V ch Ton ( o Tam ) 14. v mt s ti liu tham kho khc . 15. Ch : Nhng dng ch mu xanhcha cc ng link n cc chuyn mc h oc cc website.
mailto:[email protected]:[email protected]:[email protected]://www.dangthuchua.com/http://www.dangthuchua.com/http://www.math.vn/http://www.math.vn/http://www.math.vn/http://www.mathscope.org/http://www.mathscope.org/http://www.mathscope.org/http://www.mathscope.org/http://www.maths.vn/http://www.maths.vn/http://www.maths.vn/http://www.laisac.page.tl/http://www.laisac.page.tl/http://www.diendantoanhoc.net/http://www.diendantoanhoc.net/http://www.k2pi.violet.vn/http://www.k2pi.violet.vn/http://www.k2pi.violet.vn/http://www.nguyentatthu.violet.vn/http://www.nguyentatthu.violet.vn/http://www.nguyentatthu.violet.vn/http://dangthuchua.com/http://dangthuchua.com/http://dangthuchua.com/http://www.nguyentatthu.violet.vn/http://www.k2pi.violet.vn/http://www.diendantoanhoc.net/http://www.laisac.page.tl/http://www.maths.vn/http://www.mathscope.org/http://www.math.vn/http://www.dangthuchua.com/mailto:[email protected]5/24/2018 Bt Hsg Tinh 2010
2/46
Phn I : PHNG TRNH BPT HPT CC BI TON LIN QUAN N OHM
Phm Kim Chung THPT NG THC HA T : 0984.333.030 Mail :[email protected] Tr.
2
2
PHN I : PHNG TRNH BPT - H PTV CC BI TON LIN QUAN N O HM1. = + + +2y 2x 2 m 4xx 5Tm cc gi tr ca tham s m hm s : c cc i . S : m < -22. + =/=
=
3 21 x sin 1, xf(x)
0 , x 0
x 0Cho hm s : . Tnh o hm ca hm s ti x = 0 v chng minh hm s t cc t
ti x =0 .
3. ( )= = y f(x) | x | x 3Tm cc tr ca hm s : . S : x =0 ; x=14. Xc nh cc gi tr ca tham s m cc phng trnh sau c nghim thc :
( ) ( )+ + + =x 3 3m 4 1 x3 m4 1m 0 a) . S : 7
9
9m
7
+ =4 2x 1 x mb) . S :
5/24/2018 Bt Hsg Tinh 2010
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Phn I : PHNG TRNH BPT HPT CC BI TON LIN QUAN N OHM
Phm Kim Chung THPT NG THC HA T : 0984.333.030 Mail :[email protected] Tr.
3
3
22.Gii h PT :( ) ( )
=
=
4 4
3 3 2 2
x y 240
x 2y 3 x 4y 4 x 8y
23.Gii h phng trnh :( )
+ + = + +
=
4 3 3 2 2
3 3
x x y 9y y x y x 9x
x y x 7 . S : (x,y)=(1;2)
24. Gii h phng trnh : ( ) ( ) + + = + + =
2
2 2
4x 1 x y 3 5 2y 0
4x y 2 3 4x 7
25.Tm m h phng trnh sau c nghim : + + = + =
2 xy y x y 5
5 x 1 y m. S : m 1; 5
26.Xc nh m phng trnh sau c nghim thc : ( ) ( ) + + + =
41
x x 1 m x x x 1 1x 1
.
27.Tm m h phng trnh : ( ) + + =+ =
2
3 x 1 y m 0
x xy 1
c ba cp nghim phn bit .
28.Gii h PT :
+ + = +
+ + = +
2 y 1
2 x 1
x x 2x 2 3 1
y y 2y 2 3 1
29. ( thi HSG Tnh Ngh An nm 2008 ) .Gii h phng trnh : =
= +
x y sinxesiny
sin2x cos2y sinx cosy 1
x,y 0;4
30. Gii phng trnh : + =3 2 316x 24x 12x 3 x 31.Gii h phng trnh : ( )
( )
+ + + = +
+ + + + =
2x y y 2x 1 2x y 1
3 2
1 4 .5 2 1
y 4x ln y 2x 1 0
32.Gii phng trnh : ( )= + + +x 33 1 x log 1 2x 33. Gii phng trnh : + + = 33 2 2 32x 10x 17x 8 2x 5x x S34. Gii h phng trnh : + = +
+ + + =
5 4 10 6
2
x xy y y
4x 5 y 8 6
35. Gii h phng trnh : + + = + ++ + = + +
2 2
2 2
x 2x 22 y y 2y 1
y 2y 22 x x 2x 1
36. Gii h phng trnh :
+ =
+ = +
y x
1x y
2
1 1x y
y x
37. ( thi HSG Tnh Qung Ninh nm 2010 ) . Gii phng trnh : =
2 21 1x5x 7
( x 6)x
51
Li gii : K : >7
x5
Cch 1 : PT
+ = = +
4x 6 36(4x 6)(x 1) 0 x
2(x 1)(5x 7). x 1 5x 7
Cch 2 : Vit li phng trnh di dng : ( ) =
2 21 15x 6 x(5x 6) 1 x 1
V xt hm s : = >
2 1 5
f(t) t , t 7t 1
mailto:[email protected]:[email protected]:[email protected]://forum.mathscope.org/showthread.php?t=12343http://forum.mathscope.org/showthread.php?t=12343http://www.mediafire.com/?2h25gxl051xhttp://www.mediafire.com/?2h25gxl051xhttp://www.mediafire.com/?2h25gxl051xhttp://forum.mathscope.org/showthread.php?t=12343mailto:[email protected]5/24/2018 Bt Hsg Tinh 2010
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Phn I : PHNG TRNH BPT HPT CC BI TON LIN QUAN N OHM
Phm Kim Chung THPT NG THC HA T : 0984.333.030 Mail :[email protected] Tr.
4
4
38.( thi HSG Tnh Qung Ninh nm 2010 ) Xc nh tt c cc gi tr ca tham s m BPT sau c nghim :+ 3 2 33x 1 m( x x 1)x
HD : Nhn lin hp a v dng : ( )+ + 3
3 2x x 1 (x 3x 1) m
39.( thi HSG Tnh Qung Bnh nm 2010 ) . Gii phng trnh :+ + + = + +3 2x 3x 4x 2 (3 2) 3xx 1
HD : PT ( ) + + ++ = + +3
3(x 1) (x 1) 3x 1 3x 1 . Xt hm s : = + >3 tf t) t ,t( 0
40. ( thi HSG Tnh Hi Phng nm 2010 ) . Gii phng trnh : = + 3 23 2x 1 27x 27x 13x2 2
HD : PT = + + = 33 32x 1 (3x 1) 2(2x 1) 2 (3x 1) f( 2x 1) f(3x 1)
41.( thi Khi A nm 2010 )Gii h phng trnh : + + =+ + =
2
2 2
(4x 1)x (y 3) 5 2y 0
4x y 2 3 4x 7
HD : T pt (1) cho ta : ( ) + + = = 2
2 1].2x 5 2y 5 2y f([(2x 2x) f(1 5) 2y )
Hm s : + == + > 2 21).t f '( t) 3t f(t) (t 1 0
= = =225 4x
2x 5 2y 4x 5 2y y2
Th vo (2) ta c :
+ + =
22
2 5 4x
4x 2 3 4x 72 , vi 0 3
x 4 ( Hm ny nghch bin trn khong ) v c
nghim duy nht : =x 1
2.
42.( thi HSG Tnh Ngh An nm 2008 ) . Cho h: + =+ + +
x y 4
x 7 y 7 a(a l tham s).
Tm a h c nghim (x;y) tha mn iu kin x 9. HD : ng trc bi ton cha tham s cn lu iu kin cht ca bin khi mun quy v 1 bin kho s
= x y 0 x4 16 . t = x , t [t 3;4] v kho st tm Min .S : +a 4 2 2
43. Gii h phng trnh : + + =+ = +
4 xy 2x 4
x 3 3 y
y 4x 2 5
2 x y 2
44. Xc nh m bt phng trnh sau nghim ng vi mi x : ( ) + 2sinx sinx sinxe 1 (e 1)sinx2e e 1e 1 45.( thi HSG Tnh Tha Thin Hu nm 2003 ) . Gii PT :
+ + = 2 2
2 5 2 2 5log (x 2x 11) log (x 2x 12)
46.nh gi tr ca m phng trnh sau c nghim: ( ) ( ) + + + =4m 3 x 3 3m 4 1 x m 1 0
47.(Olympic 30-4 ln th VIII ) . Gii h phng trnh sau: +
=+
+ + = + + +
2 22
y x
2
3 2
x 1e
y 1
3log (x 2y 6) 2log (x y 2) 1
48.Cc bi ton lin quan n nh ngh a o hm : Cho +
>=
+
x
2
(x 1)e , x 0f(x)
x ax 1, x 0
. Tm a tn ti f(0) .
Cho += + + = =
2 2x x
lnx , x 0F(x) 2 4
0, , x 0
v>
= =
xlnx , x 0f(x)
0, x 0. CMR : =F'(x) f(x)
Cho f(x) xc nh trn R tha mn iu kin : >a 0bt ng thc sau lun ng x R : + | f(x a) f(x) a |. Chng minh f(x) l hm hng .
mailto:[email protected]:[email protected]:[email protected]:[email protected]5/24/2018 Bt Hsg Tinh 2010
5/46
Phn I : PHNG TRNH BPT HPT CC BI TON LIN QUAN N OHM
Phm Kim Chung THPT NG THC HA T : 0984.333.030 Mail :[email protected] Tr.
5
5
Tnh gii hn :
=
x
3
1 2
4
tanN lim
2sin
x 1
x 1 Tnh gii hn :
+=
+
2 32x 2
2 2x 0
e 1N lim
ln(1 x
x
)
Tnh gii hn :
+ + =
+ 33
x 0
3 32x x 1N
1m
xli
x Tnh gii hn :
=
sin2x
4
s
x
nx
0
ie eN lim
sinx
Tnh gii hn :
+=
0
3
5x
x 8 2
siN lim
n10x Tnh gii hn :
+=
+
2 32x 2
6 2x 0
e 1N lim
ln(1 x
x
)
Tnh gii hn : =sin2x sin3
7x
3x
0
eN lim
e
sin4x Tnh gii hn : = x 4
3x 0 38
4 xN
xim
2l
Tnh gii hn :
=
+ 9
x 0
3x 2x.3 cos4x
1 sinx 1
2N lim
sinx
Cho P(x) l a thc bc n c n nghim phn bit1 2 3 n
x x x; ; ...x . Chng minh cc ng thc sau :
a) + + + =2 n2 n
1
1
P''(x ) P''(x ) P''(x )... 0
P'(x P'( P'(x) )x)
b) + + + =2 n1 ) )
1 1 1... 0
P'(x P'(x P'(x )
Tnh cc tng sau :a) = + + +
nT osx 2cos2x ... nc(x) c osnx
b) = + + +n 2 2 n n
1 x 1 x 1 x(x) tan tan ... tan
2 2 2 2 2 2T
c) + + + = 2 3 n n 2n n n
CMR : 2.1.C 3.2.C ... n(n 1)C n(n 1).2
d) + + + += 2n
S inx 4sin2x 9sin3x ...(x) s sn innx
e) + + + = + + ++ + + + +
n 2 2 2 2 2 2
2x 1 2x 3 2x (2n 1)(x) ...
x (x 1) (x 1) (x 2) x (n 1) (x n)S
49. Cc bi ton lin quan n cc tr ca hm s : a) Cho + R: a b 0 . Chng minh rng :
+ +
n na b a b
2 2
b) Chng minh rng vi >a 3, n 2 ( n N, nchn ) th phng trnh sau v nghim :+ + ++ + + =n 2 n 1 n 2(n 1)x 3(n 2)x a 0
c) Tm tham s m hm s sau c duy nht mt cc tr : + +
= + +
22 2
2 2y (m 1) 3
x x
1 x 1 xm 4m
d) Cho n 3,n N ( n l ) . CMR : =/x 0 , ta c : + + + + +
5/24/2018 Bt Hsg Tinh 2010
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Phn II : PHNG TRNH HM V A THC
Phm Kim Chung THPT NG THC HA T : 0984.333.030 Mail :[email protected] Tr.
6
6
PHN II : PHNG TRNH HM-A THC
1. Tm hm s : f : R R tho mn ng thi cc iu kin sau :a)
=
x 0
f(x)lim 1x
b) ( ) ( ) ( )+ = + + + + 2 2f x y f x f y 2x 3xy 2y , x,y R 2. Tm hm s : f : R R tho mn iu kin sau : ( ) ( ) ( ) = + + + + 2008 2008f x f(y) f x y f f(y) y 1, x,y R 3. Tm hm s : f : R R tho mn iu kin sau : ( ) ( ) ( )( )+ = + f x cos(2009y ) f x 2009cos f y , x ,y R 4. Tm hm s : f : R R tho mn ng thi cc iu kin sau :
c) ( ) 2009xf x e d) ( ) ( ) ( )+ f x y f x .f y , x,y R
5. Tm hm s : f : R R tho mn iu kin sau : ( ) ( )+ = f y 1f x y f(x).e , x,y R 6. Tm hm s : f : R R tho mn iu kin sau : ( )( ) ( )+ = + 2f x.f x y f(y.f x ) x 7. ( thi HSG Tnh Hi Phng nm 2010 ) Tm hm f : tha mn :
( )+ + = + 2(x) 2yf(x) f(y) f y f(x) , ,x,yf R
mailto:[email protected]:[email protected]:[email protected]:[email protected]5/24/2018 Bt Hsg Tinh 2010
7/46
Phn III : BT NG THC V CC TR
Phm Kim Chung THPT NG THC HA T : 0984.333.030 Mail :[email protected] Tr.
7
7
PHN III : BT NG THCV CC TR
1. Cho + + =2 2 2a,b,c R: a b c 3 . Chng minh rng : + + 2 2 2a b b c c a 3 2. Cho cc s thc khng m a,b,c . Chng minh rng :
( ) ( ) ( ) ( ) ( ) ( ) + + 2 2 2 2 2 22 2 2 2 2 2
a b a b b c b c c a c a a b b c c a
3.
Cho cc s thc a,b,c. Chng minh rng : ( ) ( )+ + + + ++
2 2 2 2
2
a b c 81 a b 13
a b cb c a 4 42a b
4. Cho cc s thc khng m a,b,c tho mn : + + + =a b c 36abc 2 . Tm Max ca : = 7 8 9P a b c 5. Cho 3 s thc dng tu x,y,z . CMR : + +
+ + +a b c 3
a b b c c a 2
6. Cho a,b,c >0 . Tm GTNN ca : ( )+ +=6
2 3
a b cP
ab c
7. Cho cc s thc dng x,y,z tha mn : + + =2 2 2yx z 1 CMR:
+ +
2 2 22x (y z) 2y (z x) 2z (x y)
yz zx xy
8. Cho cc s thc dng a,b,c . CMR : + ++ + + + + + + +bc ca ab a b ca 3b 2c b 3c 2a c 3a 2b 6 9. Cho cc s thc dng a,b,c . CMR : + +
+ + + + + +3 3 3 3 3 31 1 1 1
abca b abc b c abc c a abc
10. Cho cc s thc tha mn iu kin : + + =+ + +2 2 2
1 1 11
a 2 b 2 c 2. CMR : + + ab bc ca 3
11. Cho cc s thc dng tha mn iu kin : + + =2 2 2ba c 3 . CMR :+ +
1 1 13
2 a 2 b 2 c
12. Cho x,y,z l 3 s thc dng ty . CMR : + + + + +x y z 3 2
x y y z z x 2
13. Cho cc s thc dng a,b,c .CMR: + + + + ++ +
2 2 2 2a b c 4(a b)a b c
b c a a b c
14. Cho cc s thc dng a,b,c tha mn : abc=1 . CMR : + + + + +3 3 3
1 1 1 3
2a (b c) b (c a) c (a b)
15. Cho 3 s thc x,y,z tha mn : xyz=1 v ( )( )( ) = /x 1 y 1 z 1 0 . CMR :
+ +
22 2x y z
1x 1 y 1 z 1
16. Cho a,b,c l cc s thc dng bt k . CMR: + + ++ + + + + + + +
2 2 2
2 2 2 2 2 2
(3a b c) (3b c a) (3c a b) 9
22a (b c) 2b (c a) 2c (a b)
17. Cho cc s thc dng a,b,c tha mn : + + =2 2 2ba c 1 . CMR :+ +
1 1 1 9
1 ab 1 bc 1 ca 2
18. Cho cc s thc a,b,c tha mn : + + =2 2 2ba c 9 .CMR : ++ +2(a b c) 10 abc 19. Cho a,b,c l cc s thc dng : a+b+c =1 . CMR : + +
3 3 3
2 2 2
a b c 1
4(1 a) (1 b) (1 c)
20.(Chn THSG QG Ng h An nm 2010 ) Cho cc s thc dng a,b,c tha mn :+ + + + + =4 4 4 2 2 2b c ) 25(9(a a b c ) 48 0 . Tm gi tr nh nht ca biu thc :
+ +=+ + +
2 2 2a b c
b 2c c 2a aF
2b
mailto:[email protected]:[email protected]:[email protected]://math.vn/showthread.php?p=38059#post38059http://math.vn/showthread.php?p=38059#post38059http://math.vn/showthread.php?p=38059#post38059http://math.vn/showthread.php?t=1107http://math.vn/showthread.php?t=1107http://math.vn/showthread.php?t=1107http://math.vn/showthread.php?t=1107http://math.vn/showthread.php?t=1107http://math.vn/showthread.php?t=1107http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107&page=2http://math.vn/showthread.php?t=1107http://math.vn/showthread.php?t=1107http://math.vn/showthread.php?p=38059#post38059mailto:[email protected]5/24/2018 Bt Hsg Tinh 2010
8/46
Phn III : BT NG THC V CC TR
Phm Kim Chung THPT NG THC HA T : 0984.333.030 Mail :[email protected] Tr.
8
8
Li gii :
T gi thit :
+ + + + + = + +
= + + + + + +
+ + + + + + +
4 4 4 2 2 2 2 2 2 4 4 4 2 2 2 2
2 2 2 2 2 2 2 2 2 2
b c ) 25(a b c ) 48 0 25(a b c ) 48 9(a b c ) 48 3(a b c )
3(a b c ) b c ) 48 0
9
3 b c
(a
1625(a a
3
Ta li c :
+ ++ + = + +
+ + + + + + + + + + +=
4 4 42 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2
a b c a b c (a b c )
b 2c c 2a a 2b a (b 2c) b (c 2a) c (a 2b) (a b b c c a) 2(a c b a cF
b)
Lic : + ++ + = + + + + + + + +2 2 2 2
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 (a b c )b b c c a a(ab) b(bc) c(ca) (a b c ) b c ca [a b a ] a b c3
Tng t :+ +
+ + + +2 2 2
2 2 2 2 2 2 a b cc b a c b) a b c .(a3
T ta c :+ +
2 2 2
F a b c
13
. Du bng xy ra khi v ch khi : a=b=c=1.
P N CA S GD&T NGH AN
p dng bt ng thc AM GM, ta c
+ ++ =
+ +
2 2 2 2 2a (b 2c)a a (b 2c)a 2a2
b 2c 9 b 2c 9 3.
Tng t+ +
+ + + +
2 2 2 2 2 2b (c 2a)b 2b c (a 2b)c 2c,
c 2a 9 3 a 2b 9 3.
Suy ra: = + ++ + +
2 2 2a b c
Fb 2c c 2a a 2b
( ) + + + + + + + 2 2 2 2 2 22 1a b c a (b 2c) b (c 2a) c (a 2b) (*)
3 9.
Li p dng AM GM, ta c
+ + + + + ++ + + + = + +
3 3 3 3 3 3 3 3 32 2 2 3 3 3a a c b b a c c b
a c b a c b a b c (**)3 3 3
.
T (*) v (**) suy ra:
( ) ( ) + + + + + +2 2 2 2 2 22 1F a b c a b c (a b c )3 9 ( ) ( ) ( ) + + + + + +2 2 2 2 2 2 2 2 22 1a b c a b c 3 a b c
3 9.
t ( )= + +2 2 2t 3 a b c , t gi thit ta c:
( ) ( ) ( )+ + = + + + + 2
2 2 2 4 4 4 2 2 225 a b c 48 9 a b c 3 a b c
( ) ( ) + + + + + + + 2
2 2 2 2 2 2 2 2 2 163 a b c 25 a b c 48 0 3 a b c3
.
Do =2 32 1
F t t f(t)9 27
vi t 3; 4 (* * *) .
M
= =t 3;4min f(t) f(3) 1 (** **) . T (***) v (****) suy ra F 1.
Vy =minF 1 xy ra khi = = =a b c 1 .
21. ( thi HSG Tnh Ngh An nm 2009 ) Cho cc s thc dng x,y,z . Chng minh rng : + +
+ + +2 2 2 2 2 21 1 1 36
x y z 9 x y y z z x
Li gii :
BT cho tng ng vi : ( )
+ + + + +
2 2 2 2 2 2 1 1 19 x y y z z x 36x y z
Ta c : ( ) + +
=
32 xy yz zx
xyz (xy)(yz)(zx)3
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9/46
Phn III : BT NG THC V CC TR
Phm Kim Chung THPT NG THC HA T : 0984.333.030 Mail :[email protected] Tr.
9
9
Do :( )+ + + +
+ + = = + ++ +
22 2
3
27 xy yz zx1 1 1 xy yz zx 27
x y z xyz xy yz zx(xy yz zx)
Li c : ( )+ + + + + + + + ++ = + +2 2 2 2 2 2 2 2 2 2 2 2y y z z x y 1 z 1) (z x 1) 29 x 6 x (y 3 (xy yz zx)
Nn :
( )
+ + + = + + + + + + + +
22 27 9VT 4 3 (xy yz zx) . 108 6 (xy yz zx)
xy yz zx xy yz zx
+ + + = + +
9108 6 2 (xy yz zx) 1296 VT 36xy yz zx
P N CA S GD&T NGH AN :
Bt ng thc cn chng minh tng ng(xy + yz + zx)(9 + x2y2+ z2y2+x2z2) 36xyz
p dng bt ng thc Csi ta c :
xy + yz + zx 3 2 2 23 x y z (1)
V 9+ x2y2+ z2y2+x2z2 4 4 412 x y z 12 hay 9 + x2y2+ z2y2+x2z2 3 xyz 12 (2)
Do cc v u dng, t (1), (2) suy ra:
(xy + yz + zx)(9 + x2y2+ z2y2+x2z2) 36xyz (pcm).Du ng thc xy ra khi v ch khi x = y = z =1
22.( thi HSG Tnh Qung Ninh nm 2010 ) Cho cc s thc dng x,y tha mn k : + + =x y 1 3xy . Tm gi tln nht ca : = +
+ + 2 23x 3y 1
My(x 1) x y 1) x
1
y(
Li gii :
Ta c : = + + + 3xy x y 1 2 xy 1 xy 1 xy 1 (*)Ta c :
( ) += +
+ +
+ += + = =
2
2 2 2 2 2 2 2 2 22
3xy 3xy 1 (1 3x1 1 1 3xy(x y) (x y)
y y (3
3x 3y 1 2xyM
y (3x 1) x (3y 1) x 9xy 3x 1) x (x y(3y 1) x y 4x) y123.( thi HS G Tnh Qung Bnh nm 2010 ) Cho cc s thc dng a, b, c . CMR :
+ + ++3 3
3 3
3
3
c a b c
b c aa
a b
b c
HD :
+ +
+ +
3 3
3 3
3 3 3
3 3 3
a a1
b b
a b c3
b c a
a3
b
24.( thi HSG Tnh Vnh Phc nm 2010 ) . Cho x, y, z 0 tha mn : + + =2 2 2yx z 1 . Tm gi tr ln nht cabiu thc : = + +P 6(y z x ) 27xyz
HD :+ + + = +
2 2 22 2 2y z 1 x6 2(y z ) x 27x. 6 2(1 x ) x 27x
2P
2 ( )=MaxP 10
25.( thi HSG Tnh Hi Phng nm 2010 ) . Cho + + =2 2 20: a bb,c ca, 1 . Chng minh rng :+ + 3 3 3
62b 3ca
7
HD : C thdng cn bng h s hoc Svacx 26. Cho x,y,z l cc s thc dng tha mn : =xyz 1 . Chng minh rng :
+ ++ + +
+ + +
4 4 3 4 4 3 4 4 3
6 6 6 6 6 6
(x (y (z
x y
y ) z ) x )12
y xzz
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10
Li gii : t = = = =2 2 2a;y b;z cx abc 1 . Bt ng thc cho tr thnh :
+ ++ + +
+ + +
3 3 3
3 3
2 2 2 2 2
3 3 3
2
3
(a (b (c
a b
b ) c ) a )12
b acc
p dng Bt ng thc AM-GM cho 4 s ta c :
( ) ( ) ( )= + + + ++ + + ++ 42 2 3 6 4 2 4 2 4 2 6 2 4 2 4 2 4 6 6 3 3(a ab ) b a b a b b b b a b ab a a a 4 ba 27. ( thi HSG Tnh ng Nai nm 2010 ). Cho a,b,c > 0 . Chng minh rng :
+ ++ +
+
+ + ++2 2 2
1 1 1 3(a b c)
a b b c c a b2( ca )
HD :
BT+ + + + + + + + + + +
+2 2 2 2 2 2(a 1b ) (b c ) 1 1 3(a b c)2 a
(c a
b b c a
)
c 2
V ch :+
+ 2
2 2 (a b)a b2
28.( thi HSG Tnh Ph Th nm 2010 ) . Cho > + + =x,y,z 0 : x y z 9 . Chng minh rng : + +
+ ++ + +
+
3 3 3 3 3 3x y z
xy 9 yz 9 zx
z x
9
y9
29.( thi chn T Ninh Bnh nm 2010 ) . Cho a,b,c l di ba cnh mttam gic c chu vi bng 4. Chng mirng : + + +
2 2 2 272
a 2abcb c 27 HD : Bi ny th chn phn t ln nht m o hm .
30.( thi HSG Tnh Bnh nh nm 2010 ) . Cho a,b,c >0 . CMR : + + + +3 3 3b c aaca abc
bb
c
HD : + + + + + += 4 2 2 2 2 4a (a b c ) (a b c)
a b cabc 3abc 27abc
VT
31. ( thi chn HSG QG Tnh Bnh nh nm 2010) . Cho x,y,z >0 tha mn : + =2 xy xz 1 . Tm gi tr nhnht ca : += +
3yz 4zS
x 5xy
x y z
32.( thi chn HSG Thi Nguyn nm 2010 ). Cho cc s thc x,y,z tha mn iu kin : + + =+ + +
1 2 31
1 x 2 y 3 zTm gi tr nh nht ca : =P xyz
33.( thi chn HSG QG tnh Bn Tre nm 2010 ) . Cho > + + =2 2 2ba,b c :a c, 0 3 . Chng minh bt ng thc :+ +
1 1 11
4 ab 4 bc 4 ca
34.( thi chn T trng HSP I H Ni 2010). Cho cc s thc dng x,y,z . Tm gi tr nh nht ca :+ +
+ += +
2 2 2
3 3 3 2 2 2
y y z z x 1xP
3xyz
x y 3(xy yz zxz )
Li gii1 :
t : = = = =x y z
a; b; c abc 1y z x
. Lc : += ++ +
+2 2 2
b c 13
3
aP
b (a b c)c a
Ta c : + ++ + = + + = + + 2(ab bc ca)
(a b c) abc(a b c) (ab)(ac) (ab)(bc) (ac)(bc)3
Li c :
+ + + = + +
+
+ +
+
2
2 2 2 2
2
2b
b
1 a 1
a b
1 b 1 a
b
c 1 1 12 ab bc ca
a b cc acc b
1 c 12
c ca
Do : + + ++ + 2
13(ab bc ca)
(ab bc ca)P ( Vi + + ab bc ca 1 )
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11
Li gii 2 :
t : = = = =z
a; b; c ay
z
x
ybc
x1 . Lc : + + + + +
+ ++
+=
+
2 2 2
2
b c 13abc 13(a b c)
c a 3(ab b
aP
c ca) (a b c)b
35.Bi ton tng t : Cho >x,y ,z 0: xyz 1 . Chng minh rng : + + + + +2 2 2
x y z 34
x y zy z x
Li gii : t : = = = 1 1 1
a; b; c abc 1x y z
.
BT cho t r thnh :+ +
+ + + ++ + + + + +
2 2 2 2
2
a b c 3abc (a b c) 9
c a b ab bc ca a b c (a b c) . Vi : + =+ 3
3a abcb c 3
36.( thi chn i tuyn H Vinh nm 2010 ) . Cho a,b,c l cc s thc thuc on [0;1]v + + =a b c 1 . Tm gtr ln nht v nh nht ca : = +
+ +
+ +2 2 21 1 1
Pa b c1 1 1
HD : Dng pp tip tuyn v Bt ng thc : + + + +
++ +2 2 2
1 1 11 ,
x y (x yx,y 0;
)x y 1
1 1 1
37.( thi chn HSG Q G tnh Lm ng ) . Cho a,b,c l cc s thc dng . Chng minh rng :+ + + + + ++
2 2 22 2 2 2 2 2b c a ab b b bc c c ca
c aba
a
Li gii :
C1 : ( THTT) Ta c : + + + + + + + +
+
+ +
2 2 2 2 2 2
b c b cc a 2(a b c) a b cc b a
a bb a c
a
Do :
= + + + = +
++
2 2 2 2 2 2b c a a2.VT 2 b a b b 2VP
c a
a
b b
ab b
b
C2 : Ta c : + + + 2 2a ab b a b c(Mincopxki)
M : + +
+= + +
Sv
2 22 22
acx
2
o
aaVT a
b
ab bab bab
ab
b c
38.( thi chn i tuyn trng Lng Th Vinh ng Nai nm 2010 ) . Cho > =a,b,c 0 : abc 1 . Chng mirng : ++ + +2 2 2ab bc c a ba c
HD : BT + + + +a b c
a b cb c a . Ch l :
=+ + 2 2 a
c 3a a c b
a b
ab c
Li gii 2 : Ta c : + + =2 2 2 2 2 2 33ab 3 (a )bab bc b c 3b
39.( Chn T HSG QG tnh Ph Th nm 2010 ). Cho >a,b,c 0 . Chng minh bt ng thc : + + +
+ +
3
3 3
2
3
2 2a b c
b c c a b
3 2
a 2
HD : + + + + + + + +
2 2
3 3
3
b c b c b c a 1 a2 3 2
a a a 2(a b c) b c3 2
40.( thi HSG Tnh Ngh An nm 2008 ) . Cho 3 s dng a,b,c thay i . Tm gi tr ln nht ca := + +
+ + +
bc ca abP .
a 3 bc b 3 ca c 3 ab
HD : t = = = =b c
x; y; zc a
axyz
b1 . Lc :
= + + = + + + + +
z x y 1 x
P 1x 3z y 3x z 3y 3 x 3z
. Li c :
+ + + += =
+ + + + + + + + ++ + +
2 2 2
2 2 22
x x (x y z) (x y z) 3
x 3z 4x 3zx (x y z) (xy yz zx) (x y z)(x y z)
3
Do : =1 3 3
P 13 4 4
. Du = xy ra khi v ch khi : x = y = z =1 .
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12
12
P N CA S GD&T :
t ( )= = = + x a ,y b ,z c;x,y,z 0; .
Khi : = + ++ + +2 2 2yz zx xy
P .x 3yz y 3zx z 3xy
Ta c = + ++ + +2 2 2
3yz 3zx 3xy3P
x 3yz y 3zx z 3xy
= + + = + + +
2 2 2
2 2 2
x y z
3 3 Qx 3yz y 3zx z 3xy
p dng bt BCS ta c
( )
+ + + + + + + +
+ + + + +
2
2 2 2
2 2 2
2 2 2
x y zx 3yz y 3zx z 3xy
x 3yz y 3zx z 3xy
Q. x y z 3xy 3yz 3zx
( )
( )
+ +
+ + + + +
2
2
x y zQ
x y z xy yz zx. Mt khc
( )+ ++ +
2x y z
xy yz zx3
Suy ra 3Q4
,do 9 33P P .4 4
Du bng xy ra khi v ch khi = =a b c. Vy gi tr nh nht ca P bng3
.4
41.( d b HSG Tnh Ngh An 2008 ) . Cho ba s dng a,b,c tho mn : + + =2 2 2a b c 1 . Tm gi tr nh nca biu thc : = + +
+ + +
2 2 2a b c
P .b c c a a b
Li gii 1: Gi s : + + +
1 1 1b c
b c c a aa
b. p dng bt ng thc Chebysev ta c :
( ) = + + + + = + + + + + + + + + + + + +
+ +
+ +
2
2
2 2
2 2
2
2
2
a b c 1 1 1 1 1 1 1 1P . ab c c a a b 3 b c c a a b 3 b c c a a b
3 3
2(a
b c
bb c) 2 (a c3 )
Li gii 2 : p dng BT Swcharz :
+ +
+ + + + += + +
+ + +
4 2 2 2 2
2 2 2 2 2 2 2 2 2
4 4a b c (a
P .a b c) b c a) c a b) b
b c )
( ( ( c ) a(b c ) c(a( ba )
Li c : + + + +
+ =
32 2 2 2 2 2 2
2 2 2a b c . b c 1 2aa(b 2(b c )
c )32 2
42.( chn i tuyn QG d thi IMO 2005 ) . Cho a,b,c >0 . CMR : + + + + +3 3 3
3 3 3a b c(a b) (b c) (c a)38
Li gii : = = ==b c a
x; y; z ; xyz 1a b c
. Bt ng thc cho tr thnh : + + + + +3 3 31 1 1 3
8(1 x) (1 y) (1 z)
p dng AM-GM ta c :( ) ( ) ( )
=+ + + +
+ + 363 3 2
11 1 13
81 x 1 x
3
8(1 x) 2 1 x
Ta cn CM bt ng thc : + + + + +2 2 21 1 1 3
4(1 x) (1 y ) (1 z)
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13
B :( ) ( )
( )+ >++ +
2 2
1 1 1x,y 0
1 xy1 x 1 y
B ny c CM bng cch bin i tng ng a v BT hin nhin : + 2 xy(x y) (1 xy)
Do :+ + + +
+ = + = =+ ++ + + + +
2
2 2 2 2
1 1 z 1 z(z 1) 1 z z 1
1 xy z 1(VT
1 z) (1 z) (1 z) z 2z 1
Gi s : = = 3z Max{x,y,z} 1 yz z zx 1 . Xt hm s :+ +
= + + +
=2 2
2 4
z z 1 z 1; f '(z) 0, z 1
z 2z 1 (z 1)f(z)
Suy ra : =3
f(f ) 1)(z4
.
43.( thi HSG Tnh H Tnh nm 2008 ) . Cho + + =0 :x yx y,z z, 1 . Tm gi tr nhnht ca :
+ ++ + +
= 1 x 1 y 1 z
1 x 1P
y 1 z
Li gii1 : ( ) +
+
22
2
1 x
1 x
x(1 x) 1 x 1 1 x 0 1 x 0
1 1 x( lun ng )
Thit lp cc BT tng t ta c : P 2
Ch : tm Max cn s dng BT ph :
+ + ++ + + +
1 x 1 y 1 x y1 , x y
1 x 1 y 1 x y
4
5v = +MaxP 1
2
3
44.( thi HSG lp 11 tnh H Tnh nm 2008 ) . Cho > + + =x,y,z 0 :x y z 1 . Chng minh bt ng thc : + + +
+ + + + + + +
1 x 1 y 1 z x y z2
y z z x x y y z x
Gii : BT
+ + + + + + + + + + + +
+
x y z x y z 3 xz xy yz2
y z z x x y y z x 2 y(y z) z(z x) x3
(x2
y)
Ta li c :( )+ +
= + + = + + + + + + + + + +
22 2 2 xz yz zxxz xy yz (xz) (xy) (yz)
VPy(y z) z(z x) x(x y ) xyz(y z) xyz(z x) xyz(x y ) 2xyz(x y z)
M :+ +
+ + = + + 2(xy yz zx)
xyz(x y z) (xy)(yz) (xz)(zy) (zx)(xy) VP3
3
2
45.( thi HSG Tnh Qung Bnh 2010 ) . Cho + + =0 :a ba b,c c, 3 . Chng minh rng : + + + + +3 3 31 1a c ab c 1b 5
46.Cho a,b,c l di 3 cnh tam gic ABC . Tm GTNN ca : = + ++ + + 2a 2b 2c
P2b 2c a 2a 2c b 2b 2a c
HD : = + + ++
2a 6a 6a
2b 2c a (a b c)(3a)(2b 2c a)
47.Cho + + =0: a ba b,c c, 1 . Tm GTLN, GTNN ca : + + += + + + ++2 2 2a 1 b 1P b c ca 1 HD .Tm GTNN : p dng BT Mincopxki ta c :
= + + = + +
+ + + + + + + + + +
2 22 2
2 2 2 1 3 3 3a 1 b 1 c 1 a a b c
2 2 2
3P a b c
2
Tm GTLN :
B : CM bt ng thc : + + + + + + + + + +2 2 21 a a 1 b b 1 1 (a b) (a b)
Bnh phng 2 v ta c : + + + + + + + + + + + + + + 2 2 2 2(1 a a 1 a b (a b) 1 a b (a b))(1 (1 ab b ) b b)a 0
48.( thi chn HSG QG tnh Hi Dng nm 2008 ) . Cho > + + =a,b,c 0:a b c 3 . Tm gi tr nh nht ca biuthc : += +
+ + +
2 2 2
3 3 3
a b c
a 2b b 2c c 2aP
HD : AM-GM ngc du .
Ta c : = = + + = + +
2 3 33 2
3 3 3 6
a 2ab 2ab 2 2 2 4a a a b a a b(a a 1) a b ab
3 9 9 9a 2b a 2b 3 ab
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Phn III : BT NG THC V CC TR
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14
Do :( )+ +
+ + + + = + + 2
a b c2 4 7 4(a b c) (ab bc ca) 1
9 9P (a b
9c
3)
3
49.( chn T trng chuyn Bn Tre ) . Cho x,y,z 0 . Tm GTLN ca : + + + + + +
= 1 1
x y z 1 (1 x)(1 y)(1M
z
Gii : t + + =x y z t 0 , ta c :+ + +
+ + +
3x y z 3
(1 x)(1 y)(1 z)3
. Lc : + + 31 2
M 7
t 1 (t 3)
Xt hm s : + +
=3
1 27, t 0
t 1 (t 3
t
)
f( )
50.Cho >a,b,c 0 . Chng minh rng : + + + + ++ + + + +
4 4 4 2 2 23a 1 3b 1 3c 1 a b c
b c c a a b 2
HD : Ta c : + = + + + =44 4 4 4 1 31 a a a a3 4 2a 1 4a
Do : = + +
3 4
Svacxo
4a 4a...
b c ab acVT
51.Cho >a,b,c 0 . Chng minh rng : + + + + ++ + + + +
1 1 1 9 4 4 4
a b c a b c a b a c b c
HD :
52. Cho > + + =a,b,c 0 :a b c 1 . Chng minh rng :( ) ( ) ( )
+ +
+ + +
b c c a 3 3
4a 3c ab b 3a b c 3b ac
b
c
a
53.Cho >a,b,c 0 . CMR : + + + + + + + + + + 2 2 2 2 2 2 2 2 2a 1 1 1 1
6 a b c3a 2b c 3b 2c a 3c 2a b
b c
54.Cho > + + =a,b,c 0:ab bc ca 3 . CMR : + + + + +2 2 2a b c
abc2a bc 2b ca 2c ab
55. Cho >a,b,c 0 . CMR : + + ++ + + + +
3 3 3
2 2 2
1 a 1 b 1 c3
1 a c 1 c b 1 b a
56.Cho > =a,b,c 0:abc 27 . CMR : + + + + +
1 1 1 3
21 a 1 b 1 c
57.Cho >a,b,c 0 . CMR : + + + + + + +
2
1 1 1 27
b(a b) c(c b) a(a c) (a b c)
58.Cho >a,b,c 0 . CMR : + + ++ + + + +b c c a a b a b c 3a b c
59.Cho (a,b,c 1;2) . CMR : + +
c b
b c
b a a c1
4b c c a 4c ab a b4a
60.Cho > =a,b,c 0 : abc 1 .CMR : ++ + + +
3 6
a b c ab bc1
ca
61.Cho >x ,y ,z 0 . CMR : + + + + + + + 2 2 2
3 3 3
x z 1 x y z
2 y z xxyz y xyz z xyz x
y x z y
62.Cho + + => 1 1 1 1a
a,b,c
b c
0: . CMR :+ +
+ +
+ + +
2 2 2a b c a b c
a bc b ac c ba 4
63.Cho >x ,y ,z 0 . Tm Min ca : = + + + + + + + +
3 3 3 3 3 33 3 3
2 2 2
x y zP 4(x y ) 4(y z ) 4(z x ) 2
y z x
64.Cho > + + =a,b,c 0: b ca 3 . CMR : + + + +a b c ab bc ca 65.Cho > =a,b,c 0:abc 1 . CMR: + +
+ + + + + +
1 1 11
a b 1 b c 1 c a 1
66.Cho >x ,y ,z 0 . CMR : + + + + + + + + + + +
x1
x (x y )(x z) y (x y )(y z) z (x z)(y z)
y z
67.( thi HSG Tnh Bnh Phc nm 2008 ). Cho >a,b,c 0 . CMR : + ++ + + + +
3 3 3
2 2 2 2 2 2
a b c a b c
2a b b c c a
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15
68.( thi HSG Tnh Thi Bnh nm 2009 ) .Cho cc s thcx , y , ztha mn + + =2 2 2x y z 3 . Tm gi tr ln nhca biu thc: = + + + + +2 2F 3x 7y 5y 5z 7z 3x
69.( thi HSG TP H Ch Minh nm 2006 ) . Cho a,b,c l cc s thc khng m tha: + + =a b c 3 . Chng minh+ +
+ + +
2 2 2
2 2 2
a b c 3
2b 1 c 1 a 1.
70. Cho a,b,c > 0 . Chng minh rng : + + + + +
2a 2b 2c3
a b b c c a
HD : t == = =a ;y ;z xb cxc a
yz 1b
. p dng B : ( )+ ++ +2 2
1 1 xy 11 xy1 x 1
2
y
71.Chng minh c c Bt ng th c :a) ( )+ + ++ + >2 2 2b c c a a blog a log logb c 3 a,b,c 2 b) ( )+ + >
+ + + + +
b c alog c log a log 9
a ,b,c 1b c c a a b a b c
2
c)72. Cho + + =0:xyx,y yzz zx, 3 . Tm gi tr nh nht ca : += + + + + 22 3 2 3 23 22P x y (xy z z 1) (y 1) (z 1)x
Gii :73.
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Phn IV : GII HN DY S
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16
PHN IV : GII HNDY S
1. Cho dy s :( )
1
2
n 1 3 n
x 1
x 7 log x 11+
=
= +. Chng minh dy s c gii hn v tnh gii hn .
HD : Xt hm s : 23
f (x(x) 17 lo 1) 5g , x (0; )+ = , ta c :2
x (0;5)11)
2xf '(x) 0
n,
(x l 3
=
+ <
Do :0 f(5) f(x ) f(0) 5< < < = > =
+
Do :2 1
) 1x f ,x .( .. .>= quy np ta c :n
x 1, n>
Li c : k k k kk k
k
2
2
k
2
k 1 2
(x 3) 1)x
1
x 2x (xx x 0
3x 3x 1+
+
<
++ < > ng vi
kx 1>
T ta c :1 2 n n 1
x ....x x x 1+> > > >> . Dy s gim v b chn di nn tn ti gii hn hu hn .
Gi s :( )2
n 2
a a 3
1
limx a 0 a a 1
3a
+=
+
> = =
+) TH3 : Nu0
0 1x< < , Xt hm s :2
2
x(xf(x)
3)
13x
+
+= trn khong (0;1) ta c :
2 2
2 2
(x 1)x (0;1) 0 f(0) f(
xf '(x) 0,
(x) f( ) 1
1)3x1
= < =
+
Do :2 1
f(x ) (0;1x ),...= quy np ta c :n
(0; nx 1),
ta c : k k k kk k
k
2
2
k
2
k 1 2
(x 3) 1)x
1
x 2x (xx x 0
3x 3x 1+
+
>
++ > < ng vi
k0 1x< <
Do :1 2 n n 1
0 x x ... x x 1+< < < < < < . Dy s tng v b chn trn nn tn ti gii hn hu hn . Gi s :
( )2n 2
a a 3
1limx a 0 a a 13a
+
= +> = =
Kt lun :n
limx 1=
10.( Bi ton tng t ) . Cho 0; a 0 > > l hai s ty . Dy 02n n nn 1 2
n
(u 3a)
a
u
{u } : uu ,n 0,1,...
3u+
=
+= =
+
. Chng minh d
c gii hn v tm gii hn .
11.( Chn i tuyn H Vinh nm 2010 ) . Cho dy s0
2n n n
n 1
n
1
1 2(u 1
u
){u }: uu , n 0,1..
u 1.+
>
+ + +
= =
. Tmn
limu
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19
12.( thi chn T HSG QG KonTum nm 2010 ) . Cho dy s thcn
{a } xc nh nh sau :1
n 1 n
n
1
1(na a
a
a+
=
= +
Chng minh rng : nn
alim 2
n+=
13.( thi HSG Tnh Hi Dng nm 2006 ) . Cho dy s thc n1 n 1
2
n
x2006; x 3x
x 1+ +==
. Tm
nxlim x+
14.(thi HSG Tnh Ph Th nm 2008 ) . Cho dy s n{x } tha mn :1
n 1 n n n n
1
x (x 1)(x 2)(x 3
x
1 , 0x ) n+
=
+ +
+ + >=
. tn
i
n
i 1
1
xy
2= += . Tm nlimy .
HD : ( )2
2 2
n 1 n n n n n n n nn n n 1
1 1 1x (x 1)(x 2)(x 3) 1 x 3x 1 x 3x 1
x 2 x 1 1x
x+
+
+ + + + = + + = + = + +
= + +
Sau chng minh dy tng v khng b chn trn .
15.Cho dy 1n 2
n 1 n n
x a 1):
2010x(
0 9xx
2 0 x+
= >
= +. Tm : 1 2 n
n2 3 n 11 1 1
x x xlim ...
x x x ++
+ + +
HD : Xt hm s :2x 2009x
f(x) , x 12010 2010
= + > . Ta c : f(x) > 0 , x 1 > f (x) f (1) 1 > = . Bng quy np chng mi
c rng :n
x 1, n> . Xt hiu :2
n n n n
n 1 n n n 1 n
x x x (xx 0,
2010 2010 201
1)x
0x 1 x x+ +
> = = > >
Gi s ( ) 2nlimx a a 1 201 2009a a 0;a 10a a = > = + = = ( Khng tha mn ). Vy nlimx = + Li c :
2 n n 1 n
n 1 n n n 1 n n n
n 1 n n 1 n n 1
x 1 12010x x ) x 1) 2010 2010
x 1
x x2009x 2
(x 1)(x 1) x 1 x 010(x x (x ++ +
+ + +
= = = =
+
16.( Bi tng t ) . Cho dy s : 124n n
n 1 n
x 1
): xx x N *
2
(x, n
4+
=
= +
. Tm gii hn23 23 23
1 2 n
2 3 n 1
x x xlim ...
x x x +
+ + +
17.( thi HSG Tnh Bnh Phc nm 2008 ) . t 2 2f(n) (n n 1) 1+ + += vi n l s nguyn dng .Xt dy sn n
f(1).f(3).f(5)...f(2n 1)(x
f(2).f(4).f(6)...f nx
)):
(2
= . Tnh gii hn ca dy s : 2
n nu n .x=
HD : Ch :2
2
f(k 1) (k 1)
f(k) (k 1
1
1)
+
+
+=
18. Chody sn
(a ) xc nh bi : n
i 1
1
2
i n
2a
a n
008
,n 1a=
= >
=
. Tnh 2n
nim al n+
HD : Ta c ( ) ( )22 2
1 2 n n n 1 n n n 1
n 1a n n 1 a n a aa ... a a
11 a
n
= = =
++ + + (1)
Trong (1) cho n=1,2,3.v nhn n l i tm : an
19.Cho dy s (
nx ) tha : 1 n 1
n
2006x 1,x 1 (n 1)1 x
+= = + +.Chng minh dy s ( nx ) c gii hn v tm gii hn y
20.( thi HSG QG nm 2009 ) . Cho dy s 1n 2
n 1 n 1 n 1
n
1x
2):
x 4x xx
2
(
, n 2
x
=
+ + =
. Chng minh rng dyn
(y ) v
n 2
n
i 1 i
1y
x== c gii hn hu hn khi n v tm gii hn .
Gii :
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20
Xt hm s :2
x xf(x)
2
4x+ += , ta c :
2
2x 4 1f '(x) 0,
24x
4xx0 >
+
+= + >
Li c :2 1 1
f(x ) 0,(do x 0)....x = > > bng quy np ta chng minh cn
x 0, n> .
Xt hiu :2 2
n 1 n 1 n 1 n 1 n 1 n 1 n 1
n n 1 n 1 n2
n 1 n 1 n 1
x 4x x x 4x x 4xx x 0,(do x
2x 0,
xn
x 4)
2 x
+ + + = = = >>
+
+
Suy ra dyn
{x } tng vn
x 0, n> .Gi s tn ti gii hn hu hnn
n 0im ( )a l x a
+
= > . Suy ra :
22a aa a a a 0
2
4a 4a+= += + = (V l ) .
Vy dyn
{x } tng v khng b chn trn nn :n
nlimx
+
= +
Li c :
( )2
2n 1 n 1 n 1 2 n n n 1 n 1
n n n 1 n 1 n 1 n n n 1 n 1 2 2 2
n 1 n n 1 n n 1 n
x 4x x x (x x ) xx 2x 4x x (x x )
1 1 1x x
x xx .x xx
2 .x x
+ + = = + = = =
Do : 1n n2 2 2
1 2 n 1 n ni 1
n
ni 1 1
1 x1 1 1 1 1 1 1y ... lim y 6
x x x x xx x x +
=
+= = + + + = =
.
21.Xt dy s thc n(x ), n N xc nh bi : 03n n 1 n 1
2009
6x 6sin(x
x
), n 1x
=
= . Chng minh dy c gii hn hu h
v tm gii hn .
HD : S dng bt ng thc :3
xx ins x,x
6x 0
Xt hm s : 3f(x) 6x 6 sin x , x 0= > . Ta c :3 2
1 6(1 cosx)f '(x) 0, x>0
3 (6x 6sin x)
= >
Do : f(x) 0 x, 0> > . M2 1 1 n n 1
f(x ) 0(do x 0) ...x f(x ) 0, nx = > > = >
Xt hiu : 33
3
n 1 n 1 n 1
n n 1 n 1 n 1 n 12 2
n 1 n 1 n 1 n 1 n 13
n 1
)x 6sin(x
6x x 6sin(xx 6x x 0)
6sin(x ) 6sin6x x 6x x(x )
= = , xt : 2n n
2
n
2
n n
n 1 n
n
n n
n
8x 11x 3x x 9x +11x + 3 x 0,
9x +11x + 3x 0
x+
+ + = > =
+>
Gi s ( )n
2
n
a 1lim x a a 0 a 9a 1a 3 3
a8
1+
+=
= > = =
+
( Khng tha mn ) n
nlim x+
= +
Do :n n
n 1
2
n n n
x 11 3lim lim 9 3
x x x+ +
+= + + =
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21
24.Cho dy sn
(u ) xc nh bi cng thc
1
2 2
n+1 n n
u = 2008
u = u - 4013u + 2007 ; n 1, n N.
a)Chng minh:n
u n + 2007; n 1, n N .
b)Dy s (x n
n
1 2 n
1 1 1x = + + ... + ; n 1, n N.
u - 2006 u - 2006 u - 2006
) c xc nh nh sau:
Tm nlimx ?
25.( thi HSG Tnh Tr Vinh-2009)Cho dy s (n
U ) xc nh bi:1
33n 1 3 n
U 1
4U log U 1 , n 1
3+
=
= + +
Tmn
n
limU+
26. Cho dy s ( )nn
0
xn n
n 1 x
x 1
): 2 l(x x 1
ln 2 1
n2 1x
2+
=
+
=
. Chng minh dy (xn
HD : Chng minh dy gim v b chn di .
) c gii hn v tm gii hn .
27. Cho phng trnh : n n 1x x .... x 1 0+ + + = . Chng t rng vi n nguyn dng th phng trnh c nghim dunht dng
nx v tm
nxlim x+
.
28.Cho dy sn
{u } xc nh bi1
n n
n 2n
1
C n
u
u .4 =
=
. . Tmn
limu
29.( thi HSG Tnh Ngh An nm 2008 ) . Cho ph ng trnh:x
1x n 0
2008 + = (1). Chng minh rng: vi m
N*phng trnh (1) c nghim duy nht, gi nghim l x n . Xt dy (xn), tm lim (xn + 1 - xnp n :
).
Vi n N*x
1x n
2008 +, xt f (x) = ; x R.
f/x
ln2008
2008(x) = - - 1 < 0 x R.
=> f(x) nghch bin trn R (1).
Ta c:n
n 1
1f(n) 0
2008
1f(n 1) 1 0
2008 +
= >
+ = f(x) =0 c nghim x n (n; n + 1) (2). T (1) v (2) => pcm.
Ta c: xnnx
1
2008- n = > 0 => xn > n.
=> 0 < xnn
1
2008- n < .
Mt khc: lim n1 02008= => lim(xn - n) = 0.
Khi lim (xn - 1 - xn) = lim{[xn + 1- (n + 1)] - (xn - n) + 1} = 1
MT S BI TON TM GII HN KHI BI T CNG THC T NG QUT C A DY S .
30.Chody s ( ) 1n n 1n
n 1
2
24
u
u : 9uu , n 2
5u 13
=
+
=
. Tmn
limu ?=
Gii :
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31.Cho dy s ( ) 1n2
n n 1
1
2
2
uu :
u u 1 , n 2
=
=
. Tm nn
ulim
n+
HD: Tm c :n 1
n
2u cos
3
=
v ch :x
n nu u1
0 lim 0n n n+
=
32.Cho dy s ( )1
n 2
n 1
n
u
u :
2 2 1 uu2
1
2
, n 2
=
=
. Tm nn
n
lim 2 .u
+
HD : Tm cn n 1
u sin2 .6
= suy ra : n
n
n n
n
n
sin3.2
lim 2 .u lim3 3
3.2
+ +
= =
33.Cho dy s ( )1
n 1nn
2
n 1
u
u :u
1 1
3
u, n
u2
+ +
=
=. Tm n
nnlim 2 .u+
HD : Tm cn n 1
u tan3.2
=
34.Cho dy s ( ) 1nn 1
n
n 1
2
3
u, n 2
2(2n 1
u :
u)u 1
u
=
=
+
. Tmi
n
ni 1
lim u+
=
35.Cho dy s : 12
n 2 n n 1
1
2
u 2u N *
u
u
u , n+ +
=
=
= +
. Tm n 1
nn
ulim
u++
HD : Tm c ( ) ( )n
n
n2u 1 1
42 2= +
. Suy ra :
( ) ( )
( ) ( )( )
n 1
n 1
n 1
n 1
x nn nn
1
2
2 2 2
12 1
1 1 1u 4lim
u 21 1 11 1
41 11
2 1
22 2
2 22
++
+ +
+
+
+ + = = = + + +
+
36.Cho dy s ( )1
n n 1
n
n 1
u
u :u
1 3
3
3 u, n 2
u
+=
=
. Tnh nn
ulim
n+
HD :n
nu tan
3
=
37.Cho dy sn
(u ) xc nh nh sau : nu 2 2 2 .... 2= + + + + ( n du cn ) . Tnh1 n
nn
2u .u ...u
lim2+
HD : t :n
n
n n 1
ux x cos
2 2 +
= = v ch : 1 2 n 1 2 nn nn 1
sinu .u ...u 1 2x ...x
2 2si
.
2
x
n+
= =
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23
38.Cho dy s 1n
2
n 1 n n n
1b
2):
1 1b b
(b
b (n 1)2 4
+
=
+
+
=
. Chng minh dy hi t v tmn
nlim b+
HD : Chng minh :n n n 1
1b .cot
2 2 +
=
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24
PHN V : HNH HC KHNG GIAN
1. Cho hnh chp tam gicu c th tch l 1. Tm gi tr ln nht ca bn knh mt cu ni tip hnh chp. 2. Cho t din ABCD c : AB=a; CD=b ; gc gia AB v CD bng .Khong cch gia AB v CD bng d. Tnh th tc
khi t din ABCD theo a,b,d v .3. Trong cc t din OABC c OA, OB, OC i mt vung gc vi nhau v th tch bng 36. Hy xc nh t din sao
cho din tch tam gic ABC nh nht.
4. Cho hnh hp ABCD.A 1B1C1D 1 . Cc im M, N di ng trn cc cnh AD v BB 11
MA NB
MD NB=sao cho . Gi I, J ln
lt l trung im cc cnh AB, C 1D 15. Gi O l tm ca mt hnh t din u . T mt im M bt k trn mt mt ca t din , ta h cc ng vung
ti ba mt cn li. Gi s K, L v N l chn cc ng vung gc ni trn. Chng minh rng ng thng OM i trng tm tam gic KLN.
.Chng minh rng ng thng MN lun ct ng thng IJ.
6. Cho hnh chp S.ABC . T im O nm trong tam gic ABC ta v cc ng thng ln l t song song vi cc cnSA, SB, SC tngng ct cc mt (SBC), (SCA), (SAB) ti cc im D,E,F .
a) Chng minh rng : OD DE DF 1SA SB SC
+ + =
b) Tm v tr ca im O trong tam gic ABC th tch ca hnh chp ODEF t gi tr ln nht. 7. Cho hnh hp ABCD.A1B1C1D 1 . Hy xc nh M thuc ng cho AC1 v im N thuc ng cho B1D 1 ca m
phng A1B1C1D 1 sao cho MN song song vi A 1
8. Cc im M, N ln lt l trung im ca cc cnh AC, SB ca t din u S.ABC . Trn cc AS v CN ta chn ccim P, Q sao cho PQ // BM . Tnh di PQ bit rng cnh ca t din bng 1. D.
9. Gi O l tm mt cu ni tip t din ABCD. Chng minh rng nu 0ODC 90= th cc mt phng (OBD) v (OADvung gc vi nhau .
10. Trong hnh chp tam gicu S.ABC (nh S ) di cc cnh y bng 6 . di ng cao SH = 15 . Qua B mt phng vung gc vi AS, mt phng ny ct SH ti O . Cc im P, Q tng ng thuc cc cnh AS v BC sao
cho PQ tip xc vi mt cu tm O bn knh bng2
5. Hy tnh di b nht ca on PQ.
11. Cho hnh lp phng ABCD.A 1B1C1D 1 cnh bng a . ng thng (d) i qua D 1 v tm O ca mt phng BCC 1Bon thng MN c trung im K thuc ng thng (d) ; M thuc mt phng (BCC1B1
12. Cho t din ABPM tho mn cc iu kin :) ; N thuc mt y (ABCD
Tnh gi tr b nht ca di on thng MN . 0 2AM BP; MAB ABP 90 ; 2AM.BP AB = = = . Chng minh rng mt
cu ng knh AB tip xc vi PM.13. ( thi HSG Tnh Qung Ninh nm 2010 ) Cho im O c nh v mt s thc a khng i . Mt hnh chp
S.ABC thay i tha mn : OA OB OC a; SA OA;SB OB;SC OC= = = ; 0 0 0ASB 90 BSC 60 CSA; ; 120= = = . Chnminh rng :
a. ABC vung . b. Khong cch SO khng thay i .
Gii :a) t : SO = x .
Ta c : Cc tam gic OAS, OBS, OCS vung nn : 2 2SA SB SC ax= = = .Do : 2 2 2 2 2AB S SB a )A 2(x= =+ ; 2 2 2 0 2 2SC 2SA.SCAC SA os120 3(x. a )c= =+ ;
2 2 2 0 2 2SB SC 2SB.SBC os6C.c a )0 (x= + = 2 2 2AB BCAC = + hay tam gic ABC vung ti B.
b) Gi M l trung im AC , do cc tam gic SAC, OAC l cc tam gic cn nn :SM AC
AC (SOM) AC OSOM AC
Tng t, gi N l trung im AB, ta CM c : AB SO Suy ra : SO (ABC) .Do mi im nm trn ng thng SO ucch u A, B, C . Suy ra SO i qua tm ngtrn ngoi tip M ca tam gic A BC . Trong cc tam gi c vun g ABC v SBO ta c h
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thc :2 2 2
2 2 2
1 1 1
BM AB BC
1
BM
1 1
OB BS
= +
=
+
2 2 2 2
1 1
OB BS
1 1
AB BC + = + 2 2
2 2 2 2 2 2 2
1 1 1 1 33a x a
22(x x a2x
) axa a + = + = =
14.( thi HSG Tnh Vnh Phc nm 2010 ) . Cho hnh chp S.ABCD cy ABCD l hnh ch nht , AB = a ;BC 2a= . Cnh bn SA vung gc vi y v SA=b . Gi M l trung im SD, N l trung im AD .
a) Chng minh AC vung gc vi mt phng (BMN) b) Gi (P) l mt phng i qua B, M v ct mt phng (SAC) theo mt ng thng vung gc vi BM
Tnh theo a, b khong cch t S n mt phng (P) . Li gii :
t AS x;AB y;AD z x.y y.z z.x 0;| x| b;| y | a;| z | a 2= = = = = = = = = Ta c : AC AD AB y z= + = +
v
1BN AN AB z y
2= =
Do :2
2 2 2(ay a 0 A1 2)
AC. C BNN z2 2
B = = =
Li do : 1MN SA MN AC2
=
Hay : AC (BMN) AC BM Gi s (P) ct (SAC) theo giao tuyn (d) BM M do (d) v AC ng phng (d)//(AC)
Gi O (AC) (BD)=
Trong mt phng (SDB) : SO ct BM ti I. Qua I k ng thng (d) // (AC) ct SA, SC ln ltti H, K . Mt phng (MHBK) l mt phng (P) cn dng . Li v : I l trng tm tam gic SDC v HK//AC nn :
SH SK SI 2
SC SA SO 3= = = (1)
Theo cng thc tnh t s th tch ta c :
SMBK SMHB
SDBA SDCB
V VSM SB SK 1 SM SH SB 1. . ; . .
V SD SB SA 3 V SD SC SB 3= = = =
2
SABCD
SKMHB SKMB SMHB SDBA
V2 b 2V
aV V V
3 3 9 = =+ = = (2)
Ta li c :KMHB MKH BKH
1 1 1S MI.HK BI.HK BM.HKS
2S
2 2= + == + (3)
M : 2 22 2 2
HK AC a 3.a
(a 2)3 3 3
+= = = ; ( ) ( )1 1
BM AM AB AS AD AB x z y2 2
= = + = +
( )
2 22 2 2 2 2 2 6aBM) z
1 1 3 b( x y b a BM
4 4 2 2
= + = +
+ =+
(4)
T (3), (4) suy ra :2 22 2
KMHB
a 3(b1 b 2 6a )6S .
2
a 3a
2 3 6
++= = (5)
T (2), (5) suy ra :2
SKMHB
2 2 2 2KMHB
b 23V 18a 2d(S,(P))
S 9a. 3(
2ab
6a 6ab ) 3(b )+ += = =
15.( thi HSG Tnh Bnh Phc nm 2010 ) . Cho hnh lp phng ABCD.ABCD c cnh bng a . Trn AB lyim M, trn CC ly im N , trn DA ly im P sao cho : AM CN D'P x x a)(0= = = .
a) CMR tam gic MNP l tam gic u, tm x din tch tam gic ny nh nht . b) Khi ax
2= hy tnh th tch khi t din BMNP v bn knh mt cu ngoi tip t din .
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16.( thi HSG Tnh B Ra Vng Tu nm 2008 ) . Cho t din ABCD c cc cnh AB=BC=CD=DA=a ,AC x; BD y= = . Gi s a khng i, xc nh t din c th tch ln nht.
17.( thi HS G Tnh B Ra Vng Tu nm 2009 ) Cho khi t din ABCD c th tch V . im M thuc min trotam gic ABC . Cc ng thng qua M son g song vi DA, DB, DC theo th t ct cc mt phng (DBC), (DCA),(DAB) tng ng ti A 1 ; B1 ; C 1
a) Chng minh rng :.
1 1 1MA MB MC
1DA DB DC
+ + =
b) Tnh gi tr ln nht ca khi t din1 1 1
MA B C khi M thay i .
18.( thi HSG Tnh Hi Phng nm 2010 ) . Cho t din OABC c OA, OB, OC i mt vung gc . Gi ; ; llt l gc to bi cc mt phng OBC, OAC, OAB vi mt phng (ABC ).
a) Chng minh rng : 2 2 2 2 2 2tan tan tan 2 tan .tan .tan + + + = b) Gi s OC=OA+OB . Chng minh rng : 0OCA OCB ACB 90+ + =
19.( thi HSG Tnh Ngh An nm 2008 ) .Cho t din ABCD c AB = CD, AC = BD, AD = BC v mt phng (CAB)vung gc vi mt phng (DAB). Chng minh rng:
1CotBCD.CotBDC = .
2
Li gii : t : BCD ; BDC= = Ta c :
BAC BDCABC DCB
ABC BCD
= = =
= =
BAD BCDCBD ADB
ABD CDB
= = =
= =
Gi H l hnh chiu ca C ln AB . t HC x= .
DoCBA DAB
CH DH(CBA) (BDA)
=
Trong tam gic vung BHC :HC HC x
sin BC ADBC sin sin
= = = =
HC x xtan BH
BH BH tan = = =
.
Trong tam gic vung AHC :HC HC x
sin AC BDAC sin sin = = = = .
HC x xtan AH
AH AH tan = = =
Trong tam gic BCD : ( ) ( )2 2
2 2 2
2 2
x x x xCD BC os 2 . cos
sin sinsiBD 2BC.BD.c
n sin= = ++ + +
(1)
Li c :2 2 2H AD 2AHD .AH oc sAD.+ =
2 22
2 2
x x x xHD 2 . .cos
tan sintan sin +
= (2)
M tam gic CHD vung nn:
2 2(1) (
22)
CH HDCD+
= + ( )2 2 2 2
2
2 2 2 2
x x x x x x x x2 . cos x 2 . .cos
sin sin tan sinsin sin tan sin+ + + = +
+
2 22 2 1(1 cot ) (1 cot ) 2(cot .cot 1) 1 cot (1 cot ) 2cot .cot cot .cot 2
+ + + + = + + + =
P N CA S GD&T :
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t AD BC a ,AC BD b, AB CD c ,BAC A ,ABC B, ACB C.= = = = = = = = =
Ta c ABC nhn v ABC = DCB = CDA = BAD.
Suy ra ( )BCD ABC B; ABD BDC CAB A, 1= = = = =
H CM AB ,v ( ) ( )CAB DAB nn ( ) ( )2 2 2CM DAB CM MD CM DM CD , 2 . + =
p dng nh l cosin cho tam gic BMD ta c ( )2 2 2MD BM BD 2BM.BD.cosMBD, 3= +
T (1), (2), (3) ta c 2 2 2 2CM BM BD 2BM.BD.cosA CD+ + = 2 2 2 2 2 2BC BD 2BM.BD.cosA CD a b 2abcosA.cosB c+ = + =
1cosC cosA.cosB sinA.sinB 2cosA.cosB cotA.cotB .
2 = = =
20.( thi HSG Tnh Ngh An nm 2008 ) .Cho khi chp S. ABCD c y ABCD l hnh bnh hnh. Gi M, N, P lnlt l trung im ca cc cnh AB, AD, SC. Chng minh rng mt phng (MNP) chia khi chp S.ABCD thnh haphn c th tch bng nhau.
21.( thi HSG Tnh Ngh An nm 2009 ) . Cho tam gic ABC , M l mt im trong tam gic ABC. Cc ngthng qua M song song vi AD, BD, CD tn g ng ct cc mt phng (BCD), (ACD) , (ABD) ln lt ti A, B, C .Tm M sao cho
MA'.MB'.MC't gi tr ln nht.
Li gii1 : tDABC MABD MAC BDC MBC AA B C
V ; ;V; V V V V V V V VV V= = = + += = v :
DA a; BD b; DC c; M A' x;MB' y ;MC ' z= = = = = =
Ta c : CV d(C,(ADB)) MC' z
V d(M,( ADB)) CD c= = = ; tng t : A B
V Vx y x y z; 1
V a V b a b c= = + + =
p dng bt ng thc AM-GM : 3x y z xyz abc
1 3 xyza b c abc 27
= + + . Du = xy ra
x y z 1
a b c 3 = = =
Do : MA'.MB'.MC' t gi tr ln nht khi v ch khi M l trn
Tm tam gic ABC .
Li gii 2 : t : DA a; BD b; DC c; M A' x ;MB' y ;MC' z= = = = = =
Ta c :A'M x
A M .DA DADA a
= =
;B M y
B M .DB .DBDB b
= =
;
P N S GD&T :
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Trong mt phng (ABC) :AM BC = {A1}; BM AC = {B1}, CM AB = {C 1}
Trong (DAA1K n g thng qua M song song vi AD ct DA
) :1 ti A
Xt tam gic DAA1 MBC1
1 ABC
SMAMA'
DA AA S
= =c MA // AD nn
Tng t ta c MAC1
1 ABC
SMBMB'DB BB S
= = , 1 MAB1 ABC
MC SMC'DC CC S
= =
Suy ra ( )MBC MAC MAB ABCMA' MB' MC'
1 doS S S SDA DB DC
+ + = + + =
Ta c 3MA ' MB' MC' MA ' MB' MC'
3 . .DA DB DC DA DB DC
+ +
Suy ra MA.MB.MC 1
27DA.DB.DC (khng i)
Vy gi tr ln nht MA.MB.MC l1
27DA.DB.DC, t c khi
1 1 1
1 1 1
MA MB MCMA' MB' MC' 1 1DA DB DC 3 AA BB CC 3= = = = = =
Hay M l trng tm tam gic ABC
22.( Tp ch THTT : T10/278 ; T10/288 ) . Cho t din S.ABC vi SA=a; SB =b ; SC = c . Mt mt phng ( ) thay i qua trng tm ca t din ct cc cnh SA, SB, SC ti cc im SA, SB, SC ti cc im D, E, F tng ng .
a) Tm gi tr nh nht ca cc biu thc :2 2 2
1 1 1
SD SE SF+ +
b) Vi k : a=b=c=1, tm gi tr ln nht ca : 1 1 1SD.SE SE.SF SF.SD
+ +
Li gii : t : SD x; SE y ; SF z= = =
G l trng tm t din nn : ( )1 1 SA 1 a
SG SA SB SC .SD .SD4 4 SD 4 x
= + + = =
Do D,E,F, G ng phng nn :a b c
4x y z
+ + = . T ta c :
( )2
2 2 2
2 2 2 2 2 2 2 2 2
1 1 1 a b c 1 1 1 16a b c (1)
x y zx y z x y z a b c16
+ + + + + + + + + +
=
Du bng xy ra
2 2 2
2 2 2
2 2 2
a b c
4a
a b c
4
x
y
z
b
a b c
4c
+ +
+ +
+ +
=
=
=
23.( thi HSG Tnh Ngh An nm 2009 ) . Cho t din ABCD c di cc cnh bng 1 . Gi M, N ln lt l truim ca BD, AC . Trn ng thng AB ly im P , trn DN ly im Q sao cho PQ song song vi CM . Tnh PQ v th tch khi AMNP . Li gii :
Gi s : AB x; AC y; AD z= = =
v :AP
m;AQ n.AC (1 n)ADAB
= = +
Ta c :1
x.y y .z z.x2
= = =
Lc :
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( )1 n
AC y; AM x z ; AP m.x; AQ n.AN (1 n )zAD .y (1 n )z2 2
= = + = = + = +
Suy ra : ( )1
CM AM AC x 2y z2
= = +
nPQ AQ AP mx y (1 n)z
2= = + +
Do CM // PQ nn :
km
2
n 2PQ kCM k k 2 3
k1 n
2
=
= = =
=
Vy : ( ) ( )221
PQ 2y x z |PQ 1 1
| 2y x z3
3PQ
9 3 3= = = =
P N CA S GD&T :
Trong mt phng (ACM) k NI // CM (I AM)Trong mt phng (BCD) k BK // CM (K CD)Trong (ABD) DI ct AB ti P Trong (AKD) DN ct AK ti Q
PQ l giao tuyn ca (DNI) v (ABK) ,do NI // CM, BK // CM nn PQ // CM
Gi E l trung im PB, ME l ng trung bnh tam gic BPD nn ME // PD hay ME // PI Mt khc t cch dng ta c I l trung im AM nn P l trung im AE.Vy AP = PE = EB
Suy ra AP 1AB 3
=
MC l ng trung bnh tam gic DBK nn BK = 2CM = 3
Suy raPQ AP 1
BK AB 3= = PQ =
1
3BK =
3
3
AMNP
AMCB
V AM AN AP 1 1 1. . .
V AM AC AB 2 3 6= = =
VAMCB1
2 = VABCD (Do M l trung im BD)
ABCD l t din u c di cnh bng 1 nn V ABCD2
12
= (vtt)
Suy ra VAMCB1 2 2
.2 12 24
== . Vy VAMNP1
6= V AMCB
2
144= (vtt)
24.( d b khi D 2008 ) . Cho t din ABCD v cc im M, N, P ln lt thuc cc cnh BC, BD, AC sao choBC 4BM; AC 3AP; BD 2BN= = = . Mt phng (MNP) ct AD ti Q . Tnh t s
AQ
ADv t s th tch hai phn ca kh
t din ABCD c phn chia bi (MNP).Li gii :
t : AB b;AC c; AD d= = =
Ta c :
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( )1
AN b d (2)2
= +
1AC 3AP AP c (3)
3= =
Do C,D,I v M, N, I thng hng nn :
AI mAC (1 m)AD 3 1
1 1c (1 m)d n b c (1 n) b d
4 4 2 2AI n AM (1
m
n)AN
= + + = + + +
=
+
n
ID 14 2AI 3AD AC2DI CD1 n IC 3
AI 2AM 3AN2 IN 2NI 2M
m
n 2
1 m 1m
2c d
N1 3
3n 1 n AI IM 30 2 2
4 2
= = = = +
=
=
= =
= =
= + + =
Gi s : AQ kAD=
. Do P, Q , I thng hng nn :
3p 1 pp
p 1 3 53 2AQ pAP (1 p)AI kd c (1 p) c d 5AQ 3AP 2AI 3PQ 2QI3 2 2 33(1 p)
kk 52
== = + = + + = + =
==
Suy ra :QI 3
PI 5=
Ta li c : IQND QPMCDN
IPMC IPMC
V VIQ IN ID 3 2 1 2 13. . . .
V IP IM IC 5 3 3 15 V 15= = = = (4)
M :( )( )
BCDABCD
PMCI MIC
d A,(BCD) .SV AC CB.CD.sinC 3 4 2 4. . .
V PC MC.CI.sinC 2 3 3 3d P,(MIC) .S= = = = (5)
T (4) v (5) suy ra : PQDNMC PQDNMC
ABCD ABMPQN
V V13 3 13 13
V 15 4 20 V 7= = =
25.( thi HSG Tnh H Tnh nm 2008) . Cho hnh chp t gic u S.ABCD c gc gia mt bn v y l . Vng cao SH ca hnh chp, gi E l im thuc SH v c khong cch ti hai mt phng (ABCD)v (SCD) bngnhau . Mt phng (P) i qua E, C, D ct SA, SB ti M, N .
a) Thit din l hnh g ?b) Gi th tch cc khi t din S.NMCD v ABCDNM ln lt l V1 , V 2 . Tm 3V2 =5V1
26.( thichn T HSG QG tnh Qung Bnh nm 2010 ) . Cho t din ABCD . Gi trung im ca AB, CD ln ll K , L . Chng minh rng bt k mt phng no i qua KL u chia khi t din ny thnh 2 phn c th tch bnhau.
.
27.( thi HS G Thnh Ph Cn Th nm 2008 ) . Trong khng gian cho hnh chp S.ABC , trng tm ABC l G .Trung im ca SG l I . Mt phng ( ) i qua I ct cc tia SA, SB, SC ln lt ti M, N, P ( Khng trng vi S ) . X
nh v tr ca mt phng ( ) th tch khi chp S .PMN l nh nht .
28.( thi HSG Tnh Hi Dng nm 2008 ) .Cho hnh lp phng1 1 1 1BAB .A CD DC cnh bng 1 . Ly cc im M
N, P, Q, R , S ln lt thuc cc cnh AD, AB, BB1 , B 1C1 , C1D 1, DD 1
29.Cho hnh chp t gic S.ABCD ,c y ABCD l mt hnh bnh hnh. Gi G l trng tm ca tam gic SAC. M l mim thay i trong min hnh bnh hnh ABCD .Tia MG ct mt bn ca hnh chp S.ABCD ti im N .
. Tm gi tr nh nht ca di ng gpkhc khp kn MNPQRSM .
t: Q =MG NG
NG MG+
a) Tm tt c cc v tr ca im M sao cho Q t gi tr nh nht .b) Tm gi tr ln nht ca Q.
30.Trong mt phng (P) cho tam gic ABC . Ly im S khng thuc (P) . Ni SA, SB, SC . I l mt im bt k trongtam gic , gi AI ct BC ti A 1 , CI ct AB ti C1 , BI ct AC ti B 1 . K IA2//SA, IB2//SB, IC2
( )2 2 2(SBC);B (SAC);A C (SAB)
//SC
. CMR : 2 2 2
1 2 1 2 1 2
SA SB SC6
A A B B C C+ +
( ) ( )3 1
BC 4BM AC AB 4 AM AB AM b c (1)4 4
= = = +
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31.( thi HSG Tnh ng Thp nm 2009 ) . Cho hnh chp S. ABCD cy ABCD l na lc gic u ni tipngtrn ng knh AD = 2a. SA vung gc vi mp ( ABCD ) v SA = a 6 .
a) Tnh khong cch t A v B n mp ( SCD ). b) Tnh din tch ca thit din ca hnh chp S.ABCD vi mp( ) song song vi mp( SAD) v cch
mp(SAD) mt khong bnga 3
4.
32.Cho t din OABC vi OA = a, OB = b, OC = c v OA, OB, OC i mt vung gc vi nhau. Tnh din tch tam gicABC theo a, b, c . Gi , , l gc gia OA, OB, OC vi mt phng ( ABC). Chng minh rng:
2 2 2
sin sin sin 1 + + = .33.Cho hai na ng thng Ax, By cho nhau v nhn AB lm on vung gc chung . Cc im M, N ln lt chuy
ng trn Ax, By sao cho AM+BN = MN . Gi O l trung im AB, H l hnh chiu ca O xung MN .a) Chng minh rng H nm trn mt ng trn c nh.
35.Khi M khc A, N khc B36.Cho hnh lp phng ABCD.ABCD c cc cnh bng a. Vi M l mt im thuc cnh AB, chn im N thuc c
DC sao cho AM+DN=aa). Chng minh ng thng MN lun i qua 1 im c nh khi M thay i.
b). Tnh th tch ca khi chp B.AMCN theo a. Xc nh v tr ca M khong cch t B ti (AMCN) t gi t
ln nht. Tnh khong cch ln nht theo a.
37. Cho hnh t din OABC a)Gi M l mt im bt k thuc min trong ca hnh t din OABC v x 1; x2; x3; x4; ln lt l khong cct M n bn mt (ABC), (OBC), (OAC) v (OAB). Gi h 1 ; h2; h3 ; h4
31 2 4
1 2 3 4
xx x x
h h h h+ + +
ln lt l chiu cao ca cc hnh chp tam gicO.ABC; A.OBC; B.OAC v C.OAB.
Chng minh tng l mt hng s .
b)Cc tia OA, OB, OC i mt hp vi nhau m 1V
Vt gc 600 . OA = a. Gc BAC bng 900
45.Cho t din ABCD c di cc cnh AB, CD ln hn 1 v di cc cnh cn li nh hn hoc bng 1. Gi H lhnh chiu ca A trn mt phng (BCD); F, K ln lt l hnh chiu ca A, B trn ng thng CD.
.
t OB+OC = m. (m >0, a > 0). Chng minh m > 2a. Tnh th tch khi t din OABC theo m v a
a) Chng minh: 2CDAF 1 -4
.
b) Tnh di cc cnh ca t din ABCD khi tch P= AH.BK.CD t gi tr ln nht.46.a) Cho hnh chp S.ABC cy ABC vung ti A , bit AB = a , A C =a 3 ; ng cao hnh chp l SA =a 3 ; M
im trn on BC sao cho BM =1
BC3
. Tnh khong cch gia hai ng thng AM v BS
b) Cho hai na ng thng Ax, By cho nhau. Hai im C, D thay i ln lt trn Ax v By sao cho:1 2 3
AC BD AB+ = .Chng minh rng: mt phng (P) cha CD v song song vi AB lun lun i qua mt im c nh
trong mt phng (Q ) cha Ax v (Q) song song By.47.( thi HSG Tnh Tr Vinh nm 2009 ) .Cho hnh chp tam gicu S.ABC c cnh y AB=a, cnh bn SA=b.
Gi M,N ln lt l trung im AB v SC. Mt mt phng ( ) thay i quay xung quanh MN ct cc cnh SA v Btheo th t P v Q khng trng vi S.
1) Chng minh rng AP bBQ a=
2) Xc nh t sAP
ASsao cho din tch MPNQ nh nht
48.Cho t din ABCD c bn knh ng trn ngoi tip cc mt u bng nhau . Chng minh rng cc cnh i dica t din u bng nhau .
49.Cho t din ABCD c cc ng cao ;BB';CC'AA' ;DD' ng quy ti mt im thuc min trong ca t din . Ccng thng ;BB';CC'AA' ;DD' li ct mt cu ngoi tip t din ABCD theo th t l
1 1 1 1;B ;C ;DA .
1 1 1 1
AA' BB' C C' DD' 8
AA BB CC DD 3+ + + .
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50.Cho t din ABCD c AB vung gc vi AC v chn ng vung gc h t A n mt phng (BCD) l trc tm tgic BCD . Chng minh rng : ( ) ( )
2 2 2 26 AB AD AB CD DB CC + ++ +
51.( thi HSG TP H Ni nm 2004 ) . Cho t din ABCD DA=a, DB=b, DC=c i mt vung gc vi nhau.Mt iM tu thuc khi t din.
a) .Gi cc gc to bi tia DM vi DA, DB, DC l , ., .CMR : 2 2 2sin sin sin 2 + + = b) .Gi
A B C DS ,S ,S ,S ln lt l din tch cc mt i din vi nh A, B, C, D ca khi t din. Tm gi
nh nht ca biu thc:A B C D
Q MA.S MB.S MC.S MD.S= + + +
52.( thi HSG TP H Ni nm 2005) .Hnh chp S.ABC c cc cnh bn i mt vung gc v SA =a, SB=b, SC=cGi A, B, C l cc im di ng ln lt thuc cc cnh SA, SB, SC nhng lun tha mn SA.SA =SB.SB=SC.SC. H l trc tm ca tam gic ABC v I l giao im ca SH vi mt phng (ABC). a) Chng minh mt phng (ABC) song song vi mt mt phng c nh v H thuc mt ng thng c nhb) Tnh IA2+IB2+IC2
53.( thi HSG TP H Ni nm 2006) .Cho t din u ABCD c cnh bng 1. Cc in M, N ln lt chuyn ngtrn cc on AB, AC sao cho mt phng (DMN) lu n vung gc vi m t phng (ABC). t AM=x, AN=y.
theo a, b, c.
a) . Cmr: mt phng (DMN) lun cha mt ng phng c nh v : x + y = 3xy. b) . Xc nh v tr ca M, N din tch ton phn t din ADMN t gi tr nh nht v ln nht.Tnh cc gi tr
54.( thi HSG TP H Ni nm 2008 ) . Cho hnh chp S.ABCD c SA lng cao v y l hnh ch nht ABCD,bit SA = a, AB = b, AD = c.
a) Trong mt phng (SBD), v qua trng tm G ca tam gic SBD mt ng thng ct cnh SB ti M vcnh SD ti N. Mt phng (AMN) ct cnh SC ca hnh chp S.ABCD ti K. Xc nh v tr ca M trn cSB sao cho th tch ca hnh chp S.AMKN t gi tr ln nht, nh nht. Tnh cc gi tr theo a, b, c
b) Trong mt phng (ABD), trn tia At l phn gic trong ca gc BAD ta chn mt im E sao cho gc bng 450
( ) ( )2 22 b c 2 b cAE
2
+ + +=. Cmr:
55. Cho hnh chp S.ABCD,y l hnh bnh hnh tm O. Hai mt bn SAB v SCD vung gc ti A v C cng hp viy gc . Bit ABC = . Chng minh SBC v SAD cng hp vi y ABCD mt gc tha mn h thc :
cotcot os.c = .56.Cho hnh chp S.ABC,y ABC l tam gic vung ti B vi AB=a, SA vung gc vi mt phng (ABC) ; mt (SAC)
hp vi mt phng (SAB) mt gc v hp vi mt phng (SBC) mt gc . Chng minh rng :
acos
cos[ ( )].SA
cos( )
+=
57.Cho hnh chp S.ABCD c y ABCD l hnh ch nht ; SA vu ng gc vi mt phng
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PHN VI : MT S KIM TRA I TUYN
S GD&T NGH ANTRNG THPT NG THC HA
Gio vin r a : Phm Kim Chung
BI KIM TRA CHT LNG I TUYN THAM GIA K THI HSG TNHNM HC 2010 2011
( Ln th 1 ) Thi gian lm bi : 180 pht_____________________________________
( )2(x 21)l x xn 1 2x
+++ =Cu 1. Gii phng trnh :
Cu 2. Xc nh tt c cc gi tr ca tham s m h phng trnh sau c nghim duy nht :2
2
22
m2x
y
m2y
y
xx
= +
= +
a ,b, c 0>Cu 3. Cho . Tm gi tr nh nht ca biu thc :4a b 3c 8c
Pa b 2c 2a b c a b 3c
+= +
+ + + + + +
( )nx ,n N *Cu 4. Cho dy s , c xc nh nh sau : 12
x3
= v nn 1
n
xx ,
2(2n 1n
)xN *
1+ +
=+
.
n 1 2 nx ..y xx .+ + += . Tm
nnlim y+
.
Cu 5. Cho hnh chp S.ABCD c SA l ng cao v y l hnh ch nht ABCD, bit SA = a, AB AD = c. Trong mt phng (SBD), v qua trng tm G ca tam gic SBD mt ng thng ct cnh ti M v ct cnh SD ti N. Mt phng (AMN) ct cnh SC ca hnh chp S.ABCD ti K. Xc nh v
ca M trn cnh SB sao cho th tch ca hnh chp S.AMKN t gi tr ln nht, nh nht. Tnh cgi tr theo a, b, c.
1 1 1 1BAB .A CD DCCu 6. Cho hnh lp phng c di bng 1 . Ly im
1E AA sao cho
1AE
3= . L
im F BC sao cho1
BF4
= . Tm khong cch t1
B n mt phng FEO ( O l tm ca hnh lp
phng ).
( ) ( )0: 0; ;f + +Cu 7. Tm hm s tho mn :
( ) ( )xf xf(y ) f f( y) x, 0, y ; )( += __________________________Ht__________________________
Thanh Chng ,ngy 03 thng 12 nm 2010
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HNG DN GII V P S
( )2(x 21)
l x xn 1 2x+
++ =Cu 1. Gii phng trnh : (1)
x 1> Li gii: iu kin :2 22(x 1)ln(x 1) x 2(x 1)ln(2x 21 0x xx ) + + = + + + =Lc : PT
( ) 2f(x) 2 x 1 ln( 2x, x 1x 1) x= + + > Xt hm s :
f ' (x) 2ln(x 1) 2x= + Ta c : ;
2 2xf ''(x) 2
x 1 x 1
= =
+ +;
2
2f '''(x) 0,
(x 1)x 1=
f ''(0) 0, f '''(0) 0= Do :
( ) 2f(x) 2 x 1 ln(x 1) 2xx= + + Vy hm s nghch bin trn khong ( )1; + . Nhn thy x 0= l mt nghim c
phng trnh (1), suy ra phng trnh c nghim duy nht x 0= .
Cu 2. Xc nh tt c cc gi tr ca tham s m h phng trnh sau :
22
22
m2x
y
m2y
y
xx
= +
= +
c nghim duy nht .
0;y 0x= =/ /Li gii : iu kin :2 2 2
2 2 2
2x y
2y
y m
x x m
= +
= +
H cho tng ng vi : (*)
x 0;y 0> >T h (*) nhn thy v tri ca cc phng trnh khng m, nn nu h c nghim (x,y) th :
2 2 2
3 2 2
x 0,y 0y x 0
(*) 2x y2x (1)
(x y)(2xy x )
y
y 0
mx m
> >= >
= =
+ + =
Do :
Do bi ton tr thnh tm tham s m phng trnh (1) c nghim dng duy nht. 3 2
f(x) 2x x , x 0 >=Xt hm s :
2
x 0
f '(x ) 6x 2x; f '(x ) 0 1x3
=
= = =
Ta c :
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2m 0Nhn vo bng bin thin ta thy, phng trnh (1) c nghim dng duy nht khi v ch khi : . Vy vi mi
m R h phng trnh cho c nghim duy nht.
a,b,c 0>Cu 3. Cho . Tm gi tr nh nht ca biu thc :
4a b 3c 8cP
a b 2c 2a b c a b 3c
+= +
+ + + + + +
Li gii :
x a b 2c a y z 2x
y 2a b c b 5x y 3z(x,y,z 0)
z a b 3c c z x
= + + = +
= + + = > = + + =
t :
( )4 y z 2x 2x y 8(z x) 4y 2x 4z 8xP 17
x y z x y x z
+ = + = + + +
Lc :
2 8 2 32 17+ 12 2 17=
( )
( )
4 3 2a t
2
2y 10 7 2b t t
2
2xR,t 0
2x2z 2c 2 1 t
+=
=
= =
=
>
Du = xy ra khi v ch khi :
( )nx ,n N *Cu 4. Cho dy s c xc nh nh sau : 12
x3
= v nn 1n
xx ,
2(2n 1n
)xN *
1+ +
=+
. t
n 1 2 nx ..y xx .+ + += . Tm
nnlim y+
Li gii :
n
n 1
n n 1 n
x 1 1x 2(2n 1)
2(2n 1)x 1 x x+
+
= = + ++ +
T : . t : nn
1v
u= , ta c : 1
n 1 n
3v
2
v 2(2n 1) v+
=
= + +
n 1
(2n 1)(2n 3)v
2+
+ +=D dng tm c cng thc tng qut ca dy :
n 1
n 1
1 1 1 1 1x
v 2n 1 2n 3 2n 1 2(n 1) 1+
+
= = + + + + +
=Do : suy ra :
n 1 2 n 1
1 1 1 1 1 1 1y x x ... 1
2 1 2.2 1 2.2 1 2.3 1 2(n 1) 1 2n 1 2n 1x ... x
= = + + + + = + + + + + + + +
+
+ +
nn n
1lim y lim 1 1
2n 1+ +
= = +
Do :
Cu 5. Cho hnh chp S.ABCD c SA lng cao v y l hnh ch nht ABCD bit SA = a, AB = b, AD = c.Trong mt phng (SBD) v qua trng tm G ca tam gic SBD mt ng thng ct cnh SB ti M v ct cnhSD ti N. Mt phng (AMN) ctcnh SC ca hnh chp S.ABCD ti K. Xc nh v tr ca M trn cnh SB sao choth tch ca hnh chp S.AMKN t gi tr ln nht nh nht. Tnh cc gi tr theo a, b, c.
Li gii : Do G l trng tm tam gic SDB, suy ra G cng l trng tm tam gic SAC. Do AG ct SC ti trung im K ca SC.
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1S 1x ; y
2
M SNx, y 1 1
SB 2SD=
=t :
SANK SAKM
SADC SACB
V VSA SN SK y SA SK SM x. . ; . .
V SA SD SC 2 V SA SC SB 2= = = =Theo cng thc tnh t s th tch ta c : Li c
SADC SACD SABC
1 1V V abc
2V
6= == v :
SANK SAKM SANKMV VV + = . Nn ta c :
SANK SAKM SANKM
SANKM
SADC SACB SABCD
V V 2V x y abc(x y)V
V V V 2 12
+ ++ = = = (*)
SM 2SN SD ySD; SM SB xSB; SG SO
S
S
SD B 3
N= = = ==
Ta li c :
1 1SO SD SB SG SN SM
3y 3x2 = + = +
V O l trung im ca BD nn : (1)
1 1 y 11 x
3y 3x 3yy
21
1
+ = =
M : M, N, G thng hng nn t (1) ta c :
2
SANKM
yabc y
3y 1 abc yV
24 8 3y 1
+ = =
Thay vo (*) suy ra :
2y 1f(y) y 1
3y 1 2
=
Xthm s :
( )
2
2
3y 2y 2f '(y) ; f '(y) 0 y
33y 1
= = =
Ta c : .
Bng bin thin :
1y4 2 1
Minf(y) y ; Maxf(y) 29 3 2
y 1
== = =
=
Nhn vo bng bin thin ta thy :
( )SANKMabc
Max V MN / /BD9
=
T ta c :
( )SANKM abcMin V8
= M l trung im SB, hoc N l trung im SD.
1 1 1 1BAB .A CD DCCu 6. Cho hnh lp phng c di bng 1 . Ly im
1E AA sao cho
1AE
3= . Ly im
F BC sao cho1
BF4
= . Tm khong cch t1
B n mt phng FEO ( O l tm ca hnh lp phng ).
1I A(0;0;0);A (0;0;1);D(1;0;0);B(0;1;0)Li gii : Chn h trc ta Ixyz sao cho
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1 1 1O ; ;
2 2 2
Lc : O l trung im AC nn ; ( )11 1
E 0;0; ;F ;1;0 ;B 0;1;13 4
1 5 3; ;
3 24OE,OF
8
=
Mt phng (OEF) i qua O v nhn vct lm vct php tuyn nn c ph ng trnh :
1 1 5 1 3 1x y z 0
3 2 24 2 8 2
=
hay : 8x 5y 9z 3 0 + =
( )12 2 2
5 9 3 11d B (OEF)
1708
;
5 9+
=
+
+=Vy :
( ) ( )0: 0; ;f + +Cu 7. Tm hm s tho mn : ( ) ( )xf xf( y) f f( y) x, 0, y ; )( +=
Li gii :
( ) ( )xf xf(1) f f(1)=Cho y = 1, suy ra : . t f(1) a= , ta c : xf(ax) f(a)= (1)
1x
a=T (1) cho , suy ra :
1f(1)=f(a) f(a)=1
a
1f(ax)
x=Cng t (1) cho ta : (2)
aax y f(y)
y
= =T (2) cho
af(y) (a 0)
y= >Th li ta thy l hm s cn tm .
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S GD&T NGH AN TRNG THPT NG THC HA
Gio vin ra : Phm Kim Chung (S>) - Nguyn Th Tha (HH)
BI KIM TRA CHT LNG I TUYN THAM GIA K THI HSG TNHNM HC 2010 2011
( Ln th 2 )Thi gian lm bi : 180 pht
_____________________________________
3 23x 3 5 2x x 3x 10x 26 0+ + + =Cu 1. Gii phng trnh:
Cu 2. Tm tt c cc gi tr ca tham s m h phng trnh sau c nghim:2 2
3 3
2 2
3 3