Phu Dao Dai So 10 CA Nam
Transcript of Phu Dao Dai So 10 CA Nam
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PHN 1 HM S BC NHT y ax b I. Kin thc c bn:
1. Hm s 0 y ax b a :
- Tp xc nh D R .- Hm s y ax b ng bin trn 0R a - Hm s y ax b nghch bin trn 0R a
- th l ng thng qua 0; , ;0b
A b Ba
.
2. Hm s hng b :- Tp xc nh D R .
- th hm s b l ng thng song song vi trc honh Ox v i qua 0;A b .
3. Hm s x :
- Tp xc nh D R .- Hm s x l hm s chn.
- Hm s ng bin trn 0; .
- Hm s nghch bin trn ;0 .
4. nh l: :d y ax b v ' : ' 'd y a x b
- d song song 'd 'a a v 'b b .
- d trng 'd 'a a v 'b b .
- d ct 'd 'a a .
Bi tp v d:
1) V th ca cc hm s sau trn cng mt h trc ta : 2x ; 2 2y x ; 3x ; 2y Hm s 2y x Hm s 2 2y x Hm s 3x
Cho 0 0x y , 0;0O cho 0 2x y , 0; 2B cho 0 3x y , 0;3D Cho 1 2x y , 1;2A cho 1 0x y , 1;0C cho 1 2x y , 1;2A
Hm s 2y l ng thng song song vi trc honh Ox v i qua im 0;2E
(Hc sinh t v hnh)2) Tm a,b th hm s ax b i qua hai im 2;1A v 1;3B .Gii: V th hm s y ax b i qua hai im 2;1A v 1;4B nn ta c h phng trnh
2 1
4
a b
a b
Gii h ta c 1a v 3b . Vy hm s cn tm l 3x .
3) Tm ta giao im ca th hai hm s bc nht: tm ta giao im (nu c) ca thhai hm s bc nht sau y 2 1x v 3 2x .Gii: Ta giao im l nghim ca h
2 1 2 1 3 2 1
3 2 3 2 1
x x x x
x y x y
.
Vy giao im cn tm l im 1;1M
4) Tm a,b ng thng ax b i qua 1;1M v song song vi ng thng 3 2y x Gii: V ng thng y ax b song song vi ng thng 3 2y x nn ta c 3a .
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V ax b i qua 1;1M nn ta c 1 1.a b , th 3a ta tm c 4b Vy ng thng cn tm l 3 4y x .
5) V th hm s cho bi nhiu cng thc:V th hm s
1, khi 1
2 , khi 1
x x y f x
x x
Vi 1x ta c 1x Vi 1x ta c 2 x Cho 1 2x y , 1;2A cho 0 2x y , 0;2C
Cho 2 3x y , 2;3B cho 1 3x y , 1;3D
BI TP
1. V th ca cc hm s sau trn cng mt h trc ta : 2 ; 2 ; 2 3 ; 2y x y x y x y .2. V th ca cc hm s sau:
a) 1, khi 02 , khi 0
x xy
x x
b) 3 1, khi 1
1, khi 1x x
yx x
c) 2 4, khi 24 2 , khi 2
x xy
x x
d)
2, khi 1
2 1, khi 1
x xy
x x
e) 1x f) 2 3y x g) 1y x h) 1 2y x
3. Tm m cc hm s:a) 1 3 y m x ng bin trn R . b) 2 3 6m x nghch bin trn R .
c) 1 3 2 y m x x m tng trn R . d) 2 3 2m x x m gim trn R .
4. Tm a,b th hm s y ax b :a) i qua hai im 1; 3A v 2;3B . c) i qua im 2; 1M v song song vi
3x
b) i qua gc ta v 2;1A . d) i qua gc ta v song song vi
2 2009y x
5. Tm m :a) th hm s 3 5x ct th hm s 2 5m x .
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b) th hm s 2 2y x song song vi th hm s 2 1 2m x m .
c) th hm s 2x trng vi th hm s 2 2y m x m .
6. Tm ta giao im nu c ca th hai ham s:a) 3 1y x v 1x b) 3 1x v 1x c) 5 6y x v 6x
7. Tm m th ca ba hm s sau ng quy (cng i qua mt im):a) 2y x v 3y x v 1 y mx
b) 1x v 3 x v 2 3 2y m x m
c) 2y x v 3 y x m v 2 5m x
8. Cho hm s 1 2 y m x a) Chng minh rng th hm s trn lun i qua mt im c nh vi mi m .
b) Tm 0m th hm s 1 2 y m x ct ,Ox Oy ti hai im ,A B sao cho OAB cn ti O.
PHN 2
Hm s bc hai - mt s dng ton lin quan
Dng 1. Kho st s bin thin v v th
Bi 1. Kho st s bin thin v v th cc hm s sau:
a)y= x2- 6x+ 3 b)y= x2- 4x+ 3 c)y= -x2 + 5x- 4
d) y= 3x2+ 7x+ 2 e) y= -x2- 2x+ 4
Bi 2. Kho st s bin thin v v th cc hm s sau:
a) 2 y x 4x 3 b) 2 y x 4 x 3 c) 2x 4 x 3
d) 2y x 4 x 3 e) 2 y x 4 x 3
Bi 3. Tm gi tr ln nht, nh nht ca hm s:
a) y = x2 -5x + 7 trn on [-2;5] b) y = -2x2 + x -3 trn on [1;3]
c) y = -3x2 - x + 4 trn on [-2;3] d) y = x2 + 3x -5 trn on [-4; -1]
Bi 4. Tm m cc bt phng trnh sau ng vi mi gi tr ca m:
a) x2 - 3x + 1 > m b) -x2 +2x - 1 > 4m c) 22x x 1 2m 1
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d) 23x x 3 3m e) x 1 x 2 x 3 x 4 m f) 2 2 x 2x 1 m m
g) x 3 x 5 x 2 x 4 3m 1
Dng 2. Lp ph-ng trnh ca parabol khi bit cc yu t ca n
Bi 5. Xc nh phng trnh cc parabol:
a) y= x2+ ax+ b i qua S(0; 1)
b) y= ax2+ x+ b i qua S(1; -1)c) y= ax2+ bx- 2 i qua S(1; 2)
d) y= ax2+ bx+ c i qua ba im A(1; -1), B(2; 3), C(-1; -3)
e) y= ax2+ bx+ c ct trc honh ti x1= 2v x2= 3, ct trc tung ti: y= 6
f) y= ax2+ bx+ c i qua hai im m(2; -7), N(-5; 0) v c trc i xng x= -2
g) y= ax2+ bx+ c t cc tiu bng6 ti x= -3 v qua im E(1; -2)
h) y= ax2+ bx+ c t cc i bng 7 ti x= 2 v qua im F(-1; -2)
i) y= ax2+ bx+ c qua S(-2; 4) v A(0; 6)
Bi 6. Tm parabol y=ax2+ bx+ 2 bit rng parabol :
a) i qua hai im A(1; 5) v B(-2; 8) b)Ct trc honh ti x1= 1 v x2= 2
c) i qua im C(1; -1) v c trc i xng x= 2 d)t cc tiu bng 3/2 ti x= -1
e) t cc i bng 3 ti x= 1
Bi 7. Tm parabol y= ax2+ 6x+ c bit rng parabol
a) i qua hai im A(1; -2) v B(-1; -10) b)Ct trc honh ti x1= -2 v x2= -4
c) i qua im C(2; 5) v c trc i xng x= 1 d)t cc tiu bng -1 ti x= -1
e) t cc i bng 2 ti x= 3Bi 8. Lp phng trnh ca (P) y = ax2 + bx + c bit (P) i qua A(-1;0) v tip xc vi ng
thng (d) y = 5x +1 ti im M c honh x = 1
Dng 3. S t-ng giao ca parabol v -ng thng
Bi 9. Tm to giao im ca cc hm s sau:
a) y= x- 1 v y= x2- 2x- 1 b) y=-x+ 3 v y= -x2- 4x +1
c) y= 2x- 5 v y=x2- 4x+ 4 d) y= 2x+ 1 v y=x2- x- 2
e) y= 3x- 2 v y= -x2- 3x+ 1 f) y= -4
1x+ 3 v y=
2
1x2+ 4x+ 3
Bi 10. Tm to giao im ca cc hm s sau:
a) y= 2x2+3x+ 2 v y= -x2+ x- 1 b) y= 4x2- 8x+ 4 v y= -2x2+ 4x- 2
c) y= 3x2+ 10x+ 7 v y= -4x2+ 3x+ 1 d)y= x2- 6x+ 8 v y= 4x2- 5x+ 3
e)y= -x2+ 6x- 9 v y= -x2+ 2x+ 3 f) y= x2- 4 v y= -x2+ 4
Bi 11 Bin lun s giao im ca ng thng (d) vi parabol (P)
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a) (d): y= mx- 1 v (P): y= x2- 3x+ 2
b) (d): y= x- 3m+ 2 v (P): y= x2- x
c) (d): y= (m- 1)x+ 3 v (P): y= -x2+ 2x+ 3
d) (d): y= 5x+ 2m+ 5 v (P): y= 5x2+ 3x- 7
Bi 12. Cho h (Pm) y = mx2 + 2(m-1)x + 3(m-1) vi m0. Hy vit phng trnh ca parabol
thuc h (Pm) tip xc vi Ox.
Bi 13Cho h (Pm) y = x2 + (2m+1)x + m21. Chng minh rng vi mi m th (Pm) lun ct
ng thng y = x ti hai im phn bit v khong cch gia hai im bng hng s.
Dng 4. Ph-ng trnh tip tuyn ca Parabol
Bi 14. Vit phng trnh tip tuyn ca (P) y = x2 - 2x +4 bit tip tuyn:
a) Tip im l M(2;4) b) Tip tuyn song song vi ng thng (d1) y = -2x + 1
c) Tip tuyn i qua im A(1:2) d) Tip tuyn vung gc vi (d2) y = 3x + 2
Bi 15. Vit phng trnh tip tuyn ca (P) y = -2x2 + 3x -1 bit tip tuyn:
a) Tip im l M(-1;3) b) Tip tuyn song song vi ng thng (d1) y = 3x -2c) Tip tuyn i qua im A(-3:2) d) Tip tuyn vung gc vi (d2) y = -3x -1
Dng 5. im c bit ca Parabol
Bi 16. Tm im c nh ca (Pm): y = mx2 + 2(m-2)x - 3m +1.
Bi 17. Tm im c nh ca (Pm): y = (m+1)x2 - 3(m+1)x - 2m -1
Bi 18. Tm im c nh ca (Pm): y = (m2 - 1)x2 - 3(m+1)x - m2 -3m + 2
Dng 6. Qu tch im
Bi 19. Tm qu tch nh ca (Pm) y = x2 - mx + m
Bi 20. Tm qu tch nh ca (Pm) y = x2 - (2m+1)x + m-1
Bi 21. Cho (P) y = x2
a) Tm qu tch cc im m t c th k c ng hai tip tuyn ti (P).
b) Tm qu tch tt c cc im m t ta c th k c hai tip tuyn ti (P) v hai tip tuyn
vung gc vi nhau.
Dng 7. Khong cch gia hai im lin quan n parabol
Bi 22. Cho (P)2x
y
4
v im M(0;-2). Gi (d) l ng thng qua M c h s gc k
a) Chng t vi mi m, (d) lun ct (P) ti hai im phn bit A v B.
b) Tm k AB ngn nht.
Bi 23. Cho (P) y = x2, ly hai im thuc (P) l A(-1;1) v B(3;9) v M l mt im thuc cung
AB. Tm to ca M din tch tam gic AMB l ln nht.
Bi 24.Cho hm s y = x2 +(2m+1)x + m2 - 1 c th (P).
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a) Chng minh rng vi mi m, th (P) lun ct ng thng y = x ti hai im phn bit v
khong cch gia hai im ny khng i.
b) Chng minh rng vi mi m, (P) lun tip xc vi mt ng thng c nh. Tm phng trnh
ng thng .
Bi 25. Cho (P) 2 y 2x x 3 . Gi A v B l hai im di ng trn (P) sao cho AB=4. Tm qu
tch trung im I ca AB.Dng 8. ng dng ca th trong gii ph-ng trnh, bpt
Bi 26. Bin lun theo m s nghim ca phng trnh:
a) x2 + 2x + 1 = m b) x2 -3x + 2 + 5m = 0 c) - x2 + 5x -6 - 3m = 0
Bi 27. Bin lun theo m s nghim ca phng trnh:
a) 2 x 5x 6 3m 1 b) 2 x 4 x 3 2m 3 c) 22x x 4m 3 0
Bi 28. Tm m phng trnh sau c nghim duy nht: 2
2 2 x 2x 4 x 2x 5 m
Bi 29. Tm m phng trnh sau c 4 nghim phn bit: 2 x x 2 4m 3
Bi 30. Tm m phng trnh sau c 3 nghim phn bit: 2 x x 2 5 2m
Bi 31. Tm gi tr ln nht, nh nht ca ( ) 4 3 2 f x x 4x x 10x 3 trn on [-1;4]
Bi 32. Cho x, y, z thay i tho mn x2 + y2 + z2 = 1. Tm gi tr ln nht v nh nht ca P= x +
y + z + xy+ yz + zx
Bi 33. Tm m bt ng thc 2 2 x 2x 1 m 0 tho mn vi mi x thuc on [1;2].
PHN III
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Ph-ng trnh bc hai & h thc VitBi tp 1 : nh gi tr ca tham s m ph-ng trnh
2 ( 1) 5 20 0 x m m x m C mt nghim x = 5 . Tm nghim kia.
Bi tp 2 : Cho ph-ng trnh2 3 0 x mx (1)
a) nh m ph-ng trnh c hai nghim phn bit.
b) Vi gi tr no ca m th ph-ng trnh (1) c mt nghim bng 1? Tm nghim kia.Bi tp 3 : Cho ph-ng trnh
2 8 5 0 x x m (1)a) nh m ph-ng trnh c hai nghim phn bit.b) Vi gi tr no ca m th ph-ng trnh (1) c mt nghim gp 3 ln nghim kia? Tm ccnghim ca ph-ng trnh trong tr-ng hp ny.Bi tp 4 : Cho ph-ng trnh
2( 4) 2 2 0m x mx m (1)
a) m = ? th (1) c nghim l x = 2 .b) m = ? th (1) c nghim kp.Bi tp 5 : Cho ph-ng trnh
2 2( 1) 4 0 x m x m (1)a) Chng minh (1) c hai nghim vi mi m.b) m =? th (1) c hai nghim tri du .c) Gi s 1 2,x x l nghim ca ph-ng trnh (1) CMR : M = 2 1 1 21 1 x x x x khng ph
thuc m.Bi tp 6 : Cho ph-ng trnh
2 2( 1) 3 0 x m x m (1)a) Chng minh (1) c nghim vi mi m.b) t M = 2 21 2x x ( 1 2,x x l nghim ca ph-ng trnh (1)). Tm min M.
Bi tp 7: Cho 3 ph-ng trnh2
2
2
1 0(1);
1 0(2);
1 0(3).
x ax b
x bx c
x cx a
Chng minh rng trong 3 ph-ng trnh t nht mt ph-ng trnh c nghim.Bi tp 8: Cho ph-ng trnh
2 2( 1) 2 0 x a x a a (1)a) Chng minh (1) c hai nghim tri duvi mi a.b) 1 2,x x l nghim ca ph-ng trnh (1) . Tm min B =
2 21 2x x .
Bi tp 9: Cho ph-ng trnh
2 2( 1) 2 5 0 x a x a (1)a) Chng minh (1) c hai nghim vi mi ab) a = ? th (1) c hai nghim 1 2,x x tho mn 1 21x x .
c) a = ? th (1) c hai nghim 1 2,x x tho mn2 21 2x x = 6.
Bi tp 10: Cho ph-ng trnh22 (2 1) 1 0 x m x m (1)
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a) m = ? th (1) c hai nghim 1 2,x x tho mn 1 23 4 11x x .b) Chng minh (1) khng c hai nghim d-ng.c) Tm h thc lin h gia 1 2,x x khng ph thuc m.Gi : Gi s (1) c hai nghim d-ng > v lBi tp 11: Cho hai ph-ng trnh
2
2
(2 ) 3 0(1)
( 3 ) 6 0(2)
x m n x m
x m n x
Tm m v n (1) v (2) t-ng -ng .Bi tp 12: Cho ph-ng trnh
2 0( 0)ax bx c a (1)iu kin cn v ph-ng trnh (1) c nghim ny gp k ln nghim kia l
2 2( 1) 0( 0)kb k ac k Bi tp 13: Cho ph-ng trnh
2 2( 4) 7 0mx m x m (1)
a) Tm m ph-ng trnh c hai nghim phn bit 1 2,x x .
b) Tm m ph-ng trnh c hai nghim 1 2,x x tho mn 1 22 0x x .c) Tm mt h thc gia 1 2,x x c lp vi m.Bi tp 14: Cho ph-ng trnh
2 2(2 3) 3 2 0 x m x m m (1)a) Chng minh rng ph-ng trnh c nghim vi mi m.b) Tm m ph-ong trnh c hai nghim i nhau .c) Tm mt h thc gia 1 2,x x c lp vi m.Bi tp 15: Cho ph-ng trnh
2( 2) 2( 4) ( 4)( 2) 0m x m x m m (1)a) Vi gi tr no ca m th ph-ng trnh (1) c nghim kp.b) Gi s ph-ng trnh c hai nghim 1 2,x x . Tm mt h thc gia 1 2,x x c lp vi m.
c) Tnh theo m biu thc 1 2
1 11 1A x x ;
d) Tm m A = 2.
Bi tp 16: Cho ph-ng trnh2 4 0 x mx (1)
a) CMR ph-ng trnh c hai nghim phn bit vi mi .
b) Tm gi tr ln nht ca biu thc 1 22 21 2
2( ) 7x xA
x x
.
c) Tm cc gi tr ca m sao cho hai nghim ca ph-ng trnh u l nghim nguyn.Bi tp 17: Vi gi tr no ca k th ph-ng trnh 2 7 0 x kx c hai nghim hn km nhau
mt n v.
Bi tp 18: Cho ph-ng trnh2 ( 2) 1 0 x m x m (1)
a) Tm m ph-ng trnh c hai nghim tri du.b) Tm m ph-ng trnh c hai nghim d-ng phn bit.
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c) Tm m ph-ng trnh c nghim m.Bi tp 19: Cho ph-ng trnh
2 ( 1) 0 x m x m (1)a) CMR ph-ng rnh (1) lun c nghim phn bit vi mi mb) Gi 1 2,x x l hai nghim ca ph-ng trnh . Tnh
2 21 2x x theo m.
c) Tm m ph-ng trnh (1) c hai nghim 1 2,x x tho mn2 21 2x x = 5.
Bi tp 20: Cho ph-ng trnh
2 2(2 1) 3 0 x m x m m (1)a) Gii ph-ng trnh (1) vi m = 3.b) Tm m ph-ng trnh c hai nghim v tch hai nghim bng 4. Tm hai nghim .Bi tp 21: Cho ph-ng trnh
2 12 0 x x m (1)Tm m ph-ng trnh c hai nghim 1 2,x x to mn
22 1x x .
Bi tp 22: Cho ph-ng trnh2( 2) 2 1 0m x mx (1)
a) Gii ph-ng trnh vi m = 2.b) Tm m ph-ng trnh c nghim.
c) Tm m ph-ng trnh c hai nghim phn bit .d) Tm m ph-ng trnh c hai nghim 1 2,x x tho mn 1 21 2 1 2 1x x .
Bi tp 23: Cho ph-ng trnh2 2( 1) 3 0 x m x m (1)
a) Gii ph-ng trnh vi m = 5.b) CMR ph-ng trnh (1) lun c hai nghim phn bit vi mi m.
c) Tnh A =3 31 2
1 1x x
theo m.
d) Tm m ph-ng trnh (1) c hai nghim i nhau.Bi tp 24: Cho ph-ng trnh
2
( 2) 2 4 0m x mx m (1)a) Tm m ph-ng trnh (1) l ph-ng trnh bc hai.
b) Gii ph-ng trnh khi m =32
.
c) Tm m ph-ng trnh (1) c hai nghim phn bit khng m.Bi tp 25: Cho ph-ng trnh
2 0 x px q (1)
a) Gii ph-ng trnh khi p = 3 3 ; q = 3 3 .b) Tm p , q ph-ng trnh (1) c hai nghim : 1 22, 1x x
c) CMR : nu (1) c hai nghim d-ng 1 2,x x th ph-ng trnh2 1 0qx px c hai nghim
d-ng 3 4,x x
d) Lp ph-ng trnh bc hai c hai nghim l 1 23 3 x v a x ; 21
1
xv
22
1
x; 1
2
x
xv 2
1
x
x
Bi tp 26: Cho ph-ng trnh2 (2 1) 0 x m x m (1)
a) CMR ph-ng trnh (1) lun c hai nghim phn bit vi mi m.b) Tm m ph-ng trnh c hai nghim tho mn : 1 2 1x x ;
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c) Tm m 2 21 2 1 26 x x x x t gi tr nh nht.Bi tp 27: Cho ph-ng trnh
2 2( 1) 2 10 0 x m x m (1)a) Gii ph-ng trnh vi m = 6.b) Tm m ph-ng trnh (1) c hai nghim 1 2,x x . Tm GTNN ca biu thc
2 21 2 1 210 A x x x x
Bi tp 28: Cho ph-ng trnh2( 1) (2 3) 2 0m x m x m (1)
a) Tm m (1) c hai nghim tri du.b) Tm m (1) c hai nghim 1 2,x x . Hy tnh nghim ny theo nghim kia.Bi tp 29: Cho ph-ng trnh
2 22( 2) ( 2 3) 0 x m x m m (1)
Tm m (1) c hai nghim 1 2,x x phn bit tho mn1 2
1 2
1 1
5
x x
x x
Bi tp 30: Cho ph-ng trnh2 0 x mx n c 3 2m = 16n.
CMR hai nghim ca ph-ng trnh , c mt nghim gp ba ln nghim kia.Bi tp 31 : Gi 1 2,x x l cc nghim ca ph-ng trnh
22 3 5 0x x . Khng gii ph-ng trnh ,
hy tnh : a)1 2
1 1x x
; b) 21 2( )x x ;
c) 3 31 2
x x d) 1 2x x
Bi tp 32 : Lp ph-ng trnh bc hai c cc nghim bng :
a) 3 v 2 3 ; b) 2 3 v 2 + 3 .Bi tp 33 : CMR tn ti mt ph-ng trnh c cc h s hu t nhn mt trong cc nghim l :
a)3 5
3 5
; b)
2 3
2 3
; c) 2 3
Bi tp 33 : Lp ph-ng trnh bc hai c cc nghim bng:
a) Bnh ph-ng ca cc nghim ca ph-ng trnh 2 2 1 0x x ;b) Nghch o ca cc nghim ca ph-ng trnh 2 2 0 x mx Bi tp 34 : Xc nh cc s m v n sao cho cc nghim ca ph-ng trnh
2 0 x mx n cng l m v n.Bi tp 35: Cho ph-ng trnh
2 32 ( 1) 0 x mx m (1)a) Gii ph-ng trnh (1) khi m = 1.b) Xc nh m ph-ng trnh (1) c hai nghim phn bit , trong mt nghim bng bnhphung nghim cn li.
Bi tp 36: Cho ph-ng trnh22 5 1 0x x (1)
Tnh 1 2 2 1 x x x x ( Vi 1 2,x x l hai nghim ca ph-ng trnh)
Bi tp 37: Cho ph-ng trnh2(2 1) 2 1 0m x mx (1)
a)Xc nh m ph-ng trnh c nghim thuc khong ( 1; 0 ).b) Xc nh m ph-ng trnh c hai nghim 1 2,x x tho mn
2 21 2 1x x
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Bi tp 38 :Cho phng trnh x2 - (2k - 1)x +2k -2 = 0 (k l tham s).
Chng minh rng phng trnh lun lun c nghim.
Bi tp 39:Tm cc gi r ca a ptrnh:
032)3( 222 axaxaa Nhn x=2 l nghim .Tm nghim cn li ca ptrnh?
Bi tp 40 Xc nh gi tr ca m trong ph-ng trnh bc hai :2 8 0 x x m
4 + 3 l nghim ca ph-ng trnh . Vi m va tm -c , ph-ng trnh cho cn mtnghim na . Tm nghim cn li y?
Bi tp 41: Cho ph-ng trnh : 2 2( 1) 4 0 x m x m (1) , (m l tham s).
1) Gii ph-ng trnh (1) vi m = 5.2) Chng minh rng ph-ng trnh (1) lun c hai nghim 1 2,x x phn bit mi m.
3) Tm m 1 2x x t gi tr nh nht ( 1 2,x x l hai nghim ca ph-ng trnh (1) ni trong phn 2/ ) .
Bi tp 42:
Cho phng trnh
1. Gii phng trnh khi b= -3 v c=22. Tm b,c phng trnh cho c hai nghim phn bit v tch ca chng bng 1
Bi tp 43:Cho phng trnh x2 2mx + m2 m + 1 = 0 vi m l tham s v x l n s.
a) Gii phng trnh vi m = 1.b) Tm m phng trnh c hai nghim phn bit x1 ,x2.
c) Vi iu kin ca cu b hy tm m biu thc A = x1 x2 - x1 - x2 t gi tr nh nht.Bi tp 44:
Cho ph-ng trnh ( n x) : x4 2mx2 + m2 3 = 01) Gii ph-ng trnh vi m = 3 2) Tm m ph-ng trnh c ng 3 nghim phn bit
Bi tp 45: Cho ph-ng trnh ( n x) : x2 2mx + m22
1= 0 (1)
1) Tm m ph-ng trnh (1) c nghim v cc nghim ca ptrnh c gi tr tuyt i bng nhau2) Tm m ph-ng trnh (1) c nghim v cc nghim y l s o ca 2 cnh gc vung ca mttam gic vung c cnh huyn bng 3.
Bi tp 46: Lp ph-ng trnh bc hai vi h s nguyn c hai nghim l:
53
41
x v
53
42
x
1) Tnh : P =44
53
4
53
4
Bi tp 47: Tm m ph-ng trnh: 0122 mxxx c ng hai nghim phn bit.
http://hocmai.vn/filter/tex/displaytex.php?x%5E2+%2B+bx+%2B+c+%3D+0http://hocmai.vn/filter/tex/displaytex.php?x%5E2+%2B+bx+%2B+c+%3D+0http://hocmai.vn/filter/tex/displaytex.php?x%5E2+%2B+bx+%2B+c+%3D+0 -
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Bi tp 48: Cho hai ph-ng trnh sau :2
2
(2 3) 6 0
2 5 0
x m x
x x m
( x l n , m l tham s )
Tm m hai ph-ng trnh cho c ng mt nghim chung.
Bi tp 49:
Cho ph-ng trnh : 2 22( 1) 1 0 x m x m vi x l n , m l tham s cho tr-c
1) Gii ph-ng trnh cho kho m = 0.2) Tm m ph-ng trnh cho c hai nghim d-ng 1 2,x x phn bit tho mn iu kin2 21 2 4 2x x
Bi tp 50: Cho ph-ng trnh :
22 1 2 3 0m x m x m ( x l n ; m l tham s ).
1) Gii ph-ng trnh khi m = 9
2
2) CMR ph-ng trnh cho c nghim vi mi m.3) Tm tt c cc gi tr ca m sao cho ph-ng trnh c hai nghim phn bit v nghim ny gpba ln nghim kia.
Bi tp 52: Cho ph-ng trnh x2 + x 1 = 0 .a) Chng minh rng ph-ng trnh c hai nghim tri du .
b) Gi 1x l nghim m ca ph-ng trnh . Hy tnh gi tr biu thc :81 1 110 13 P x x x
Bi tp 53: Cho ph-ng trnh vi n s thc x:x2 2(m 2 ) x + m 2 =0. (1)
Tm m ph-ng trnh (1) c nghim kp. Tnh nghim kp .
Bi tp 54:Cho ph-ng trnh : x2 + 2(m1) x +2m 5 =0. (1)a) CMR ph-ng trnh (1) lun c 2 nghim phn bit vi mi m.b) Tm m 2 nghim 1 2,x x ca (1) tho mn:
2 21 2 14x x .
Bi tp 55:a) Cho a = 11 6 2 , 11 6 2b . CMR a, ,b l hai nghim ca ph-ng trnh bc hai vi h snguyn.
b) Cho 3 36 3 10, 6 3 10c d . CMR 2,c d l hai nghim ca ph-ng trnh bc hai vi h snguyn.Bi tp 56: Cho ph-ng trnh bc hai :
2 22( 1) 1 0 x m x m m (x l n, m l tham s).1) Tm tt c cc gi tr ca tham s m ph-ng trnh c 2 nghim phn bit u m.2) Tm tt c cc gi tr ca tham s m ph-ng trnh c 2 nghim 1 2,x x tho mn : 1 2 3x x .
3) Tm tt c cc gi tr ca tham s m tp gi tr ca hm sy= 2 22( 1) 1 x m x m m cha on 2;3 .
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Bi tp 57:Cho ph-ng trnh :x2 2(m1) x +2m 3 =0.
a) Tm m ph-ng trnh c 2 nghim tri du.b) Tm m ph-ng trnh c 2 nghim ny bng bnh ph-ng nghim kia.
Bi tp 58: Cho ph-ng trnh : 2 26 6 0. x x a a 1) Vi gi tr no ca a th ph-ng trnh c nghim.2) Gi s
1 2,x x l nghim ca ph-ng trnh ny. Hy tm gi tr ca a sao cho 3
2 1 18 x x x
Bi tp 59: Cho ph-ng trnh :mx2 5x ( m + 5) = 0 (1) trong m l tham s, x l n.
a) Gii ph-ng trnh khi m = 5.b) Chng t rng ph-ng trnh (1) lun c nghim vi mi m.c) Trong tr-ng hp ph-ng trnh (1) c hai nghim phn bit 1 2,x x , hy tnh theo m gi tr ca
biu thc B = 2 21 2 1 210 3( ) x x x x . Tm m B = 0.
Bi tp 60:a) Cho ph-ng trnh : 2 22 1 0 x mx m ( m l tham s ,x l n s). Tm tt c cc gi tr nguynca m ph-ng trnh c hai nghim 1 2,x x tho mn iu kin 1 22000 2007x x
b) Cho a, b, c, d R . CMR t nht mt trong 4 ph-ng trnh sau c nghim2
2
2
2
2 0;
2 0;
2 0;
2 0;
ax bx c
bx cx d
cx dx a
dx ax b
Bi tp 61:1) Cho a, b , c, l cc s d-ng tho mn ng thc 2 2 2a b ab c . CMR ph-ng trnh
2
2 ( )( ) 0 x x a c b c c hai nghim phn bit.Cho ph-ng trnh 2 0 x x p c hai nghim d-ng 1 2,x x . Xc nh gi tr ca p khi
4 4 5 51 2 1 2 x x x x
t gi tr ln nht.Bi tp 62: Cho ph-ng trnh :(m + 1 ) x2 ( 2m + 3 ) x +2 = 0 , vi m l tham s.
a) Gii ph-ng trnh vi m = 1.b) Tm m ph-ng trnh c hai nghim phn bit sao cho nghim ny gp 4 ln nghim kia.
Bi tp 63: Cho ph-ng trnh: 2 23 2 2 10 4 0 x y xy x y (1)
1) Tm nghim ( x ; y ) ca ph-ng trnh ( 1 ) tho mn 2 2 10x y 2) Tm nghim nguyn ca ph-ng trnh (1).Bi tp 64: Gi s hai ph-ng trnh bc hai n x :
21 1 1 0a x b x c v
22 2 2 0a x b x c
C nghim chung. CMR
: 2
1 2 2 1 1 2 2 1 1 2 2 1 .a c a c a b a b b c b c
Bi tp 65: Cho ph-ng trnh bc hai n x :
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2 22( 1) 2 3 1 0 x m x m m a) Chng minh ph-ng trnh c nghim khi v ch khi 0 1m
b) Gi 1 2,x x l nghim ca ph-ng trnh , chng minh : 1 2 1 29
8x x x x
Bi tp 66: Cho ph-ng trnh bc hai n x :2 22 2 2 0 x mx m
a) Xc nh m ph-ng trnh c hai nghim.b) Gi
1 2,x x l nghim ca ph-ng trnh , tm gi tr ln nht ca biu thc :
1 2 1 22 4 A x x x x .
Bi tp 67: Cho ph-ng trnh bc hai n x :2( 1) 2( 1) 3 0m x m x m vi m 1. (1)
a) CMR (1) lun c hai nghim phn bit vi mi m.b) Gi 1 2,x x l nghim ca ph-ng trnh (1) , tm m 1 2 0x x v 1 22x x Bi tp 68: Cho a , b , c l di 3 cnh ca 1 tam gic . CMR ph-ng trnh
2 ( ) 0 x a b c x ab bc ac v nghim .Bi tp 69: Cho cc ph-ng trnh bc hai n x :
2
2
0(1);
0(2).
ax bx c
cx dx a
Bit rng (1) c cc nghim m v n, (2) c cc nghim p v q. CMR :2 2 2 2
4m n p q .Bi tp 70: Cho cc ph-ng trnh bc hai n x :
2 0 x bx c c cc nghim 1 2,x x ; ph-ng trnh2 2 0 x b x bc c cc nghim 3 4,x x .
Bit 3 1 4 2 1 x x x x . Xc nh b, c.Bi tp 71 : Gii cc ph-ng trnh saua) 3x4 5x2 +2 = 0b) x6 7x2 +6 = 0c) (x2 +x +2)2 12 (x2 +x +2) +35 = 0d) (x2 + 3x +2)(x2+7x +12)=24
e) 3x2+ 3x = xx 2 +1
f) (x + x1
) 4 ( )1
xx +6 =0
g) 121 2 xx
h) 20204 xx
i) (1048
3 2
2
x
x)
4
3 x
x
Bi tp 72. gii cc ph-ng trnh sau.
a) x2 5 x 5 =0 b) 5 .x2 2 x +1=0
c) ( 1 03)13()3 2 x d)5x4 7x2 +2 = 0 e) (x2 +2x
+1)2 12 (x2 +2x +1) +35 = 0 f) (x2 4x +3)(x212x +35)=16 g) 2x2+ 2x = xx 2
+1 .Bi tp 73.Cho ph-ng trnh bc hai 4x25x+1=0 (*) c hai nghim l x 1 , x 2 .
1/ khng gii ph-ng trnh tnh gi tr ca cc biu thc sau:
22
21
11
xxA ; B
22
22
1
1 44
x
x
x
x
; 52
51 xxC ;
72
71 xxD
2/ lp ph-ng trnh bc hai c cc nghim bng:a) u = 2x1 3, v = 2x23
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b) u =1x
1
1 , v =
1x
1
2 .
Bi tp 74 . Cho hai ph-ng trnh : x2 mx +3 = 0 v x2 x +m+2= 0 .a) Tm m ph-ng trnh c nghim chung.b) Tm m hai ph-ng trnh t-ng -ng.Bi tp 75. Cho ph-ng trnh (a3)x2 2(a1)x +a5 = 0 .a) tm a ph-ng trnh c hai nghim phn bit x1, x2.
b) Tm a sao cho 1x
1
+ 2x
1
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a) Gii PT khi m = 2b) C/mr phg-ng trnh cho c hai nghim tri du vi mi GT ca mc) Gi hai nghim cu PT cho l x1 ; x2 .Tm m hai nghim tho mn
3 3
1 2
2 1
x x
x x
t GTLN
Bi tp 88: Cho Ph-ng trnh : x2 mx m 1 = 0 (*)a) C/mr PT (*) c nghim x1 ; x2 vi mi GT ca m ; tnh nghim kp ( nu c ) ca PT v GT m t-ng -ng.b) t A = x12 + x22 6x1.x21) Chng minh A = m2 8m + 82) Tm m sao cho A= 83) Tm GTNN ca a v GT m t-ng ng .Bi tp 89: Cho ph-ng trnh x2 2(a 1) x + 2a 5 = 0 (1)a) C/mr PT(1) c nghim vi mi ab) Vi gi tr no ca a th (1) c nghim x1 ,x2 tho mn x1 < 1 < x2c) Vi gi tr no ca a th ph-ng trnh (1) c hai nghim x1, x2 tho mn
x12 + x2
2 =6Bi tp 90: Cho ph-ng trnh : x2 2(m+1)x + m 4 = 0 ( *)a) Chng minh (*) c hai nghim vi mi mb) Tm gi tr ca m PT (*) c hai nghim tri duc) Gi s x1 ; x2 l nghim ca PT (*)
Chn minh rng : M = (1 x1) x2 + (1 x2)x1Bi tp 91: Cho ph-ng trnh : x2 (1 2n) x + n 5 = 0a) Gii PT khi m = 0b) Chng minh rng PT c nghim vi mi gi tr ca nc) Gi x1; x2 l hai nghim cu PT choChng minh rng biu thc : x1(1 + x2) + x2(1 +x1)Bi tp 92: Cc nghim ca ph-ng trnhx2 + ax + b + 1 = 0 (b khc 1) l nhng s nguynChng minh rng a2 + b2 l hp s
Bi tp 93: Cho a,b,c l ba cnh ca tam gic .C/m:x2 + ( a + b + c) x + ab + bc + ca = 0
v nghim
Bi tp 94: Cho cc ph-ng trnh ax2 + bx + c = 0 ( a.c 0) v cx2 + dx + a = 0 c cc nghim x1; x2 v y1 ;y2 t-ng -ng C/m x1
2 + x22 + y1
2 + y22 4
Bi tp 95: Cho cc ph-ng trnh x2+ bx +c =0 (1) v x2 +cx +b = 0 (2)
Trong 2
111
cb
Bi tp 96: Cho p,q l hai s d-ng .Gi x1 ; x2 l hai nghim ca ph-ng trnhpx2 + x +q = 0 v x3 ; x4 l nghim ca ph-ng trnh qx
2 + x + p = 0
C/m : 1 2 3 4. . 2 x x x x
Bi tp 97: Cho a,b,c l ba s thc bt k .Chng minh rng t nht mt trong ba ph-ng trnh sau c nghim
:2 2 21 0; 1 0; 1 0 x ax b x bx c x cx a
Bi tp 98: Cho ph-ng trnh bc hai :x2 + (m+2) x + 2m = 0 (1)a) C/m ph-ng trnh lun lun c nnghimb) Gi x1; x2 l hai nghim ca ph-ng trnh . Tm m 2(x1
2 + x22 ) = 5x1x2
Bi tp 99: Cho ph-ng trnh x2 + a1x + b1 = 0 (1) ;x2 + a2x + b2 = 0 (2)
C cc h s tho mn 1 2 1 22a a b b .Cmr t nht mt trong hai ph-ng trnh trn c nghim
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Bi tp 100: Chng minh rng ph-ng trnh : 2 2 2 2 2 2 0a x b a c x b
V nghimNu a + b > c v a b c
Bi tp 101: Cho hai ph-ng trnh :x2 + mx + 1 = 0 (1) x2 + x + m = 0 (2)
a) Tm m hai ph-ng trnh trn c t nht mt nghim chung
b) Tm m hai ph-ng trnh trn t-ng -ng
Bi tp 102: Cho ph-ng trnh:x2 2( a + b +c) x + 3( ab + bc+ ca) = 0 (1)a) C/mr ph-ng trnh (1) lun c nghimTrong tr-ng hp ph-ng trnh (1) c nghim kp xc nh a,b,c .Bit a2 + b2 + c2 = 14Bi tp 103: Chng minh rng nu ph-ng trnh :x2 + ax + b = 0 v x2 + cx + d = 0 c nghim chung th : (b
d)2 + (a c)(ad bc) = 0Bi tp 104: Cho ph-ng trnh ax2 + bx + c = 0 .C/mr nu b > a + c th ph-ng trnh lun c 2 nghim phnbitBi tp 105: G/s x1 , x2 l hai nghim ca hai ph-ng trnh x
2 + ax + bc = 0 v x2 , x3 l hai nghim caph-ng trnh x2 + bx + ac = 0 ( vi bc khc ac ) . Chng minh x1, x3 l nghim ca ph-ng trnh x
2 + cx + ab =
0 .Bi tp 106: Cho ph-ng trnh x2 + px + q = 0 (1) .Tm p,q v cc nghim ca ph-ng trnh (1) bit rng khithm 1 vo cc nghim ca n chng ch thnh nghim ca ph-ng trnh : x2 p2x + pq = 0Bi tp 107: Chng minh rng ph-ng trnh :
(x a) (x b) + (xc) (x b) + (xc) (x a) = 0Lun c nghim vi mi a,b,c.
Bi tp 108: Gi x1; x2 l hai nghim ca ph-ng trnh : 2x2 + 2(m +1) x + m2 +4m + 3 = 0
Tm GTLN ca biu thc A = 1 2 1 22 2 x x x x
Bi tp 109: Cho a 0 .G/s x1 ; x2 l nghim ca ph-ng trnh2
2
10
2 x ax
a
Chng minh rng : 4 41 2 2 2x x
Bi tp 110 Cho ph-ng trnh 22
1 0 x axa
.Gi x1 ; x2 l hai nghim ca ph-ng trnh
Tm GTNN ca E = 4 41 2x x
Bi tp 111: Cho pt x2 + 2(a + 3) x + 4( a + 3) = 0a) Vi gi tr no ca a th ph-ng trnh c nghim kpb) Xc nh a ph-ng trnh c hai nghim ln hn 1
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H ph-ng trnh bc nht hai n
Dng
''' cybxa
cbyax
1. Gii h phng trnh
1)
3)12(412)12(
yxyx 2)
53
1
7
3
1
3
2
5
3
yx
yx
2. Gii v bin lun h phng trnh
1)
55
55
myx
ymx2)
mmyxm
myxm
3)1(
72)5(
3. Tm gi tr ca tham s h phng trnh c v s nghim
1)
23)12(
3)12(
mmyxm
mymmx2)
mnmynx
nmnymx
2
22
4. Tm m hai ng thng sau song songmy
mxmyx
1)1(,046
5. Tm m hai ng thng sau ct nhau trn Oymymxmmyx 3)32(,2
H gm mt ph-ng trnh bc nht vmt ph-ng trnh bc hai hai n
Dng
)2(
)1(22 khygxeydxycx
cbyax
PP gii: Rt xhoc y (1) ri th vo (2).1. Gii h phng trnh
1)
423
53222 yyx
yx2)
5)(3
0143
yxxy
yx
3)
100121052
13222 yxyxyx
yx
2. Gii v bin lun h phng trnh
1)
22
1222 yx
ymx2)
22
1222 yx
ymx
3. Tm m ng thng 0)1(88 mymx
ct parabol 02 2 xyx ti hai im phn bit.
H ph-ng trnh i xng loi I
Dng
0),(
0),(
2
1
yxf
yxf; vi ),( yxfi = ),( xyfi .
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PP gii: t PSPxy
Syx4; 2
1. Gii h phng trnh
1)
7
522 xyyx
xyyx2)
30
1122 xyyx
xyyx
3)
931
19
2244
22
yxyx
xyyx 4)
243
2111
33 yx
yx
5)
491
1)(
51
1)(
2222
yxyx
xyyx
6)
2
5
1722
y
x
y
x
yx
2. Tm m h phng trnh c nghim
1)
myx
yx
66
22 12)
mxyyx
yxyx
)1)(1(
8)22
3. Cho h phng trnh
32
22 xyyx
myx
Gi s yx; l mt nghim ca h. Tm m biu thc F= xyyx 22 t max, t min
H ph-ng trnh i xng loi II
Dng
0),(
0),(
xyf
yxf
PP gii: h tng ng
0),(),(
0),(
xyfyxf
yxf
hay
0),(),(0),(),(
xyfyxf
xyfyxf
1. Gii h phng trnh
1)
yxx
xyy
43
432
2
2)
yxyx
xxyy
3
32
2
3)
yxyx
xyxy
40
4023
23
4)
yxx
xyy
83
833
3
2. Tm m h phng trnh c nghim duy nht.
1)
myxx
myxy
2)(
2)(
2
2
2)
myyyx
mxxxy
232
232
4
4
H ph-ng trnh ng cp (cp 2)
Dng
)2(''''
)1(22
22
dycxybxa
dcybxyax
PP gii: t tx nu 0x 1. Gii h phng trnh
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1)
932
22222
22
yxyx
yxyx2)
42
133222
22
yxyx
yxyx
3)
16
1724322
22
yx
yxyx4)
137
152
22
xyy
yx
2. Tm m h phng trnh c nghim
1)
myxyx
yxyx
1732
1123
22
22
2)
myxyx
yxyx
22
22
54
132
Mt s H ph-ng trnh khc1. Gii h phng trnh
1)
7
122 yxyx
yx2)
180
4922 xyyx
xyyx
3)
7
2)(33 yx
yxxy4)
0)(9)(8
01233 yxyx
xy
5)
21
122
yx
yx6)
yxyx
xyxy
10)(
3)(2
22
22
2. Gii h phng trnh
1)
12
527
yxyx
yxyx3)
7
142
222
zyx
yxz
zyx
2)
523
53
2323 22
yx
xxyy
3. Tm m hai phng trnh sau c nghim chung
a) mx 31 v 12422
mx b) 01)2()1( 2 xmxm v
0122 mxx 4. Tm m h phng trnh c nghim
02
)1(
xyyx
xyayx
11
1
xy
myx
4. Tm m, n h phng trnh sau c nhiuhn 5 nghim phn bit
myxyyxmx
ynxyx
22
22
)(
1
PHN 5BT NG THC
Dng nh ngha
Chng minh cc bt ng thc sau
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1.Cho a,b,c,d > 0
a) nu a < b tha
b b tha
b >a + c
b + c
c) 1 b > 0 v c ab . Chng minh rngc + ac2 + a2
c + bc2 + b2
5.Cho a + b + c 0. Chng minh rng :a3 + b3 + c3 3abc
a + b + c 0
5.Cho ba s dng a ,b ,c ,chng minh rng :1
a3 + b3 + abc +1
b3 + c3 + abc +1
c3 + a3 + abc 1
abc
4.Cho cc s a,b,c,d tho a b c d 0. Chng minh rng :
a) a2 b2 + c2 (a b + c)2 b) a2 b2 + c2 d2 (a b + c d)2
5.a) Cho a.b 1,Chng minh rng :1
1 + a2 +1
1 + b2 2
1 + ab
a) Cho a 1, b 1 .Chng minh rng :1
1 + a3 +1
1 + b3 +1
1 + c3 3
1 + abc
b) Cho hai s x ,y tho x + y 0.Chng minh rng :1
1 + 4x +1
1 + 4y 2
1 + 2x+y
6. a,b,c,d chng minh rnga) a2 + b2 + c2 + d2 (a + c)2 + (b + d)2
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b) 1 a3 + b3 + c3d) a3(b2 c2) + b3(c2 a2) + c3(a2 b2) < 0e) (a + b + c)2 9bc vi a b cf) (a + b c)(b + c a)(c + a b) abc8. Cho hai s a ,b tho a + b 2 ,chng minh rng : a4 + b4 a3 + b39.Cho a ,b ,c 0 , chng minh rng :a) a3 + b3 + c3 3abc
b) a3b + b3c + c3a a2bc + b2ca + c2abc) a3(b2 c2) + b3(c2 a2) + c3(a2 b2) < 010. Cho a ,b ,c l di 3 cnh mt tam gic,vi a b c
Chng minh rng : (a + b + c)2 9bc
*.Cho tam gic ABC,chng minh rng :aA + bB + cC
a + b + c 3
*.Cho a ,b ,c [0;2] . Chng minh rng : 2(a + b + c) (ab + bc + ca) 4. Chng minh rng :
11.2 +
12.3 +
13.4 + +
1n(n + 1) < 1 n N
. Chng minh rng :12! +
23! +
34! + +
n 1n! < 1 n N n 2
*.Cho ba s dng a ,b ,c tho mn: ab + bc + ca = 1 . Chng minh rng :
3 a + b + c 1
abc
.11.Cho 3 s a, b, c tho mn a + b + c = 3. Chng minh rng :a) a2 + b2 + c2 3
b) a4 + b4 + c4 a3 + b3 + c3
Bt ng thc Cauchy1.Cho hai s a 0 , b 0 Chng minh rng :
a)a
b +ba 2 a , b > 0 b) a
2b +1
b 2a b > 0 c)2a2 + 14a2 + 1
1
d) a3 + b3 ab(a + b) e) a4 + a3b + ab + b2 4a2bf) (a + b)(1 + ab) 4ab g) (1 + a)(1 + b) (1 + ab )2
h)a2
a4 + 1 12 i)
1a +
1b
4a + b j)
1a +
1b +
1c
2a + b +
2b + c +
2c + a
j) (1 + a)(1 + b) (1 + ab )2 h)a2 + 2a2 + 1
2 k)a6 + b9
4 3a2b3 16
l) a
2
+ 6a2 + 2 4 m) a
2
b2 +b
2
c2 + c
2
a2 ac + cb +ba
2.Cho a > 0 , chng minh rng : (1 + a)2
1a2 +
2a + 1 16
3. Cho 3 s a ,b ,c > 0 ty . Chng minh rng:
a) a2b +1
b 2a b) a + b + c 12 ( a
2b + b2c + c2a +1a +
1b +
1c )
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4.Cho 0 < a < b , chng minh rng: a b v ab = 1 ,chng minh rng :a2 + b2
a b 2 2
Chng minh rng 12
(a + b)(1 ab)(1 + a2)(1 + b2)
12
13 .a) Chng minh rng nu b > 0 , c > 0 th :b + c
bc 4
b + c
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b)S dng kt qu trn chng minh rng nu a ,b ,c l ba s khng m c tnga + b + c = 1 th b + c 16abc
14.Cho a + b = 1,Chng minh rng: (1 +1a )(1+
1b ) 9
15.Cho a,b,c > 0 v a + b + c = 1 . Chng minh rng :
a) (1 +1a )(1+
1b )(1+
1c ) 64 b) (a + b)(b + c)(c + a)abc
8729
16.Cho 4 s a ,b ,c ,d > 0 tho mn1
1 + a+
1
1 + b+
1
1 + c+
1
1 + d 3
Chng minh rng abcd 181
17.Cho a,b,c l di ba cnh ca mt tam gic ,chng minh rng :a) ab + bc + ca < a2 + b2 + c2 < 2(ab + bc + ca)
b) abc (a + b c)(b + c a)(c + a b)
c) (p a)(p b)(p c) abc8 d)
1p a +
1p b +
1p c 2(
1a +
1b +
1c )
e) p < p a + p b + p c < 3p18.Cho 3 s a ,b ,c 0 ,tho mn a.b.c = 1
Chng minh rng : (1 + a)(1 + b)(1 + c) 819. Cho 3 s x, y, z tho mn: x2 + y2 + z2 = 1. Chng minh rng
1 x + y + z + xy + yz + zx 1 + 320 .Cho n s dng a1 ,a2 ,.,an. Chng minh rng
a)a1a2
+a2a3
+ +ana1
n b) (a1 + a2 + + an)(1a1
+1a2
+ +1an
) n2
c) (1 + a1)(1 + a2)(1 + an) 2n vi a1.a2.an = 1
21.Cho n s a1 ,a2 ,.,an [0;1] ,chng minh rng :(1 + a1 + a2 + + an)
2 4(a12 + a2
2 + + an2)
22.Cho a > b > 0 , chng minh rng : a +1
b(a b) 3 .Khi no xy ra du =
23. Cho hai s a 0 ; b 0 . Chng minh rng :
a) 2 a + 33
b 55
ab b)17125
ab17b12a5 c)
a6 + b9
4 3a2b3 16
24. Chng minh rng 1.3.5.(2n 1) < nn25.Cho ba s khng m a ,b ,c chng minh rng :
a + b + c knm nmkknm mknknm knm cbacbacba 26 .Cho 2n s dng a1 ,a2 ,.,an v b1 ,b2 ,.,bn.
Chng minh rng :n
a1.a2....an +n
b1.b2....bn n
(a1 + b1)(a2 + b2).(an + bn)
27. Chng minh rng :4
(a + 1)(b + 4)(c 2)(d 3)
a + b + c + d
1
4
a 1 , b 4 , c 2 ,d > 328*. n N chng minh rng :
a) 1.122
.133
. 144
.. 1nn ( 1 +1n )
n
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30*.Cho x1,x2,xn > 0 v x1 + x2 + .+ xn = 1 Chng minh rng
(1 +1x1
)(1+1x2
)(1+1xn
) (n + 1)n
31*.Cho cc s x1,x2 ,y1,y2,z1,z2 tho mn x1.x2 > 0 ; x1.z1 y12 ; x2.z2 y2
2Chng minh rng : (x1 + x2)(z1 + z2) (y1 + y2)
2
32*.Cho 3 s a ,b ,c (0;1). Chng minh rng trong 3 bt ng thc sau phi c mt bt ng thc sai:a(1 b) > 1/4 (1) ; b(1 c) > 1/4 (2) ; c(1 a) > 1/4 (3)
33*.Cho 3 s a,b,c > 0. Chng minh rng :
2 aa3 + b2 + 2 bb3 + c2 + 2 cc3 + a2 1a2 + 1b2 + 1c2
34** Cho x ,y ,z [0;1] ,chng minh rng : (2x + 2y + 2z)(2 x + 2 y + 2 z) 818
(HBK 78 trang 181,BT Trn c Huyn)35*.Cho a , b , c > 1. Chng minh rng :
a) log2a + log2b 2 log2
a + b2
b) 2
logbaa + b +
logcbb + c +
logacc + a
9a + b + c
36*Cho a ,b ,c > 0,chng minh rng :
a) ab + c + bc + a + ca + b 32 b) a
2
b + c + b
2
c + a + c
2
a + b a + b + c2
c)a + b
c +b + c
a +c + ab 6 d)
a3
b +b3
c +c3
a ab + bc + ca
e) (a + b + c)(a2 + b2 + c2) 9abc f)bc
a +acb +
abc a + b + c
g)a2
b + c +b2
c + a +c2
a + b a + b + c
2 ab
a + b +bc
b + c +ca
c + a
37.Cho ba s a ,b ,c tu . Chng minh rng :a2(1 + b2) + b2(1 + c2) + c2(1 +ab2) 6abc
38*Cho a ,b ,c > 0 tho :1a +
1c =
2b . Chng minh rng :
a + b2a b +
c + b2c b 4
39*Cho 3 s a, b, c tho a + b + c 1. Chng minh rng :
a)1a +
1b +
1c 9 b)
1a2 + 2bc +
1b2 + 2ac +
1c2 + 2ab 9
40*Cho a ,b ,c > 0 tho a + b + c k. Chng minh rng :
(1 +1a )(1 +
1b )(1 +
1c ) (1 +
3k)
3
41*Cho ba s a ,b ,c 0. Chng minh rng :a2
b2 +b2
c2 +c2
a2 a
b +bc +
ca
42*Cho tam gic ABC,Chng minh rng :
a) ha + hb + hc 9r b)a ba + b +
b cb + c +
c ac + a 0a) x2 + 4y2 + 3z2 + 14 > 2x + 12y + 6z
b) 5x2 + 3y2 + 4xy 2x + 8y + 9 0c) 3y2 + x2 + 2xy + 2x + 6y + 3 0d) x2y4 + 2(x2 + 2)y2 + 4xy + x2 4xy3e) (x + y)2 xy + 1 3 (x + y)
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f) 3
x2
y2 +y2
x2 8
xy +
yx + 10 0
g) (xy + yz + zx)2 3xyz(x + y + z)2.Cho 4 s a ,b ,c ,d tho b< c < d chng minh rng :
(a + b + c + d)2 > 8(ac + bd)3. Chng minh rng : (1 + 2x + 3x)2 < 3 + 3.4x + 32x+14. Cho ax + by xy , x,y > 0. Chng minh rng : ab 1/4
*5. Cho 1 x 1
2v
5
6< y 0
6** Cho a3 > 36 v abc = 1.Xt tam thc f(x) = x2 ax 3bc +a2
3
a) Chng minh rng : f(x) > 0 x
b) Chng minh rng:a2
3 + b2 + c2 > ab + bc + ca
Cho hai s x , y tho mn: x y . Chng minh rng x3 3x y3 3y + 4.Tm Gi tr nh nht ca cc hm s :
a) y = x2 +4x2
b) y = x + 2 +1
x + 2
vi x > 2
c) y = x +1
x 1 vi x > 1
d) y =x3 +
1x + 2 vi x > 2
e) y =x2 + x + 1
x vi x > 0
f) y =4x +
91 x vi x (0;1)
.Tm gi tr ln nht ca cc hm s sau:y = x(2 x) 0 x 2
y = (2x 3)(5 2x)
3
2 x
5
2
y = (3x 2)(1 x)23 x 1
y = (2x 1)(4 3x)12 x
43
y = 4x3 x4 vi x [0;4].Trong mt phng ta Oxy,trn cc tia Ox v Oy ln lt ly cc im A v B thay i sao cho ngthng AB lun lun tip xc vi ng trn tm O bn knh R = 1. Xc nh ta ca A v B onAB c di nh nht*.Cho a 3 ; b 4 ; c 2 .Tm gi tr ln nht ca biu thc
A =ab c 2 + bc a 3 + ca b 4
abc
*Tm gi tr ln nht v gi tr nh nht ca hm s y = x 1 + 5 x
PHN 5
1. BT PHNG TRNH V H BT PHNG TRNH BC NHT MT N
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Bi 1. Gii cc bt phng trnh sau
2
3 1 2 1 2.
2 3 4
. (2 1)( 3) 3 1 ( 1)( 3) 5
x x xa
b x x x x x x
2
2 2
. 8 3
3. 1 2( 3) 5 4
2
c x x
d x x x
Bi 2 Gii cc h bpt sau:
a.3 1 2 7
4 3 2 19
x x
x x
b.
56 4 7
78 3
2 52
x x
xx
2. DU CA NH THC BC NHTBi 1. Xt du cc biu thc sau:
. ( ) (2 1)( 3)
4 3. ( )
3 1 2
a f x x x
b f xx x
2
. ( ) (3 2)( 2)( 3)
. ( ) 9 1
c f x x x x
d f x x
Bi 2 Gii cc bpt sau:
2
3 7.
2 2 11 1
.2 ( 2)
ax x
bx x
2
2
1 2 3.
3 2
3 3. 1
4
c x x x
x xd
x
2
3. 1
2
2. 1
1 2
ex
x xf x
x
Bi 3. Gii cc bpt sau:. 2 1 3 5
. 3 1 2
. 3 2 2
a x x
b x x
c x x
.2 1 1d x x
3.DU CA TAM THC BC HAIBi 1. Xt du cc biu thc sau:
2 2
2 2
2 2
. ( ) 3 2 . ( ) 2 5 2
. ( ) 9 24 16 . ( ) 3 5
. ( ) 2 4 15 . 4 4
a f x x x b f x x x
c f x x x d f x x x
e f x x x f f x x x
Bi 2. Lp bng xt du cc biu thc sau:2
2 2
2 2
2 2
2
. ( ) (3 10 3)(4 5)
. ( ) (3 4 )(2 1)
. ( ) (4 1)( 8 3)(2 9)
(3 )(3 ). ( )4 3
a f x x x x
b f x x x x x
c f x x x x x
x x xd f x
x x
7.BT PHNG TRNH BC HAIBi 1. Gii cc bpt sau:
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22 2
2 2
1 3.4 1 0 .
4 3 4
. 3 4 0 . 6 0
a x x c x x x
b x x d x x
8.MT S PHNG TRNH V BT PHNG TRNH QUY V BC HAIGii cc bt phng trnh sau:
1)2 2 5
34
x xx
x
2)2 3 1
2
x xx
x
3)3 47 4 47
3 1 2 1
x x
x x
4)9
42
x
x
5)
3 4
3 2
1 2 60
7 2
x x x
x x
6)
24 2 4 2 x x x 7) 2 7 10 0x x
8) 2 23 2 5 6 0 x x x x 9)2 3
01 2
x x
x
10)
1
2+
2
2
3 2 1
4 3 3
x x x
x x x
11)2
2
2 3 4 15
1 1 1
x x x x
x x x
12)
2
2 1 42 2 2 x x x
13)
2 3
1 2 2 31 1 1
x
x x x x
14)4 3 2
2
3 20
30
x x x
x x
15)
3 23 30
2 x x x
x x
16)
4 2
2
4 30
8 15
x x
x x
17)
3 4 5
21 2 3 6 0
7 x x x x
x x
18) 2
4211
x xx x
19) 22
2151
1x x
x x
Gii h bt phng trnh sau:
2)
2 31
12 2 4
01
x
x
x x
x
3)2 12 0
2 1 0
x x
x
4)
2
2
3 10 3 0
6 16 0
x x
x x
5)2
2
4 7 0
2 1 0
x x
x x
6)
2
2
5 0
6 1 0
x x
x x
7)
2
2
3 8 3 0
17 7 6 0
x x
x x
8)
2
2
2
4 3 02 10 0
2 5 3 0
x x
x x
x x
9)2
2
2 74 1
1
x x
x
10)
2
2
1 2 21
13 5 7
x x
x x
11)2
2
10 3 21 1
3 2
x x
x x
12)
2
2
2
3 40
3
2 0
x x
x
x x
13)
2
2
4 2
2
30
12 0
2 0
4 5 0
x x
x
x
x x
x x
Phng trnh v bt phng trnh c cha tr tuyt i:
1) 2 5 4 4 x x x 2) 2 22 8 1 x x x 3) 2 5 1 1 0x x
4) 31 1 x x x 5) 2 1 2 0x x 6) 1 4 2 1x x
7) 2 23 2 2 x x x x 8) 2 5 7 4x x 9)2
2
41
2x x
x x
10)2
2
5 41
4
x x
x
11)
2 51 0
3x
x
12)
2
23
5 6
x
x x
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13)2
2x x
x
14) 2
2
21x
x 15)
2
2
4 31
5
x x
x x
16) 2 3 3x x 17)
2 1 12
2
x x
x x
18) 2 4 2 x x x
19) 3 1 2x x 20)2
2
2 41
2
x x
x x
21) 1 3 x x x x
22)2 6
22
x xx
x
23) 2 1 5x x 24) 1 2 x x x
Phng trnh v bt phng trnh c cha cn :
1) 2 2 4 2 x x x 2) 23 9 1 2 x x x 3) 2 12 7 x x x
4) 221 4 3 x x x 5) 21 2 3 5 0 x x x 6) 2 1
2 12
xx
x
7)
2 16 533 3
xxx x
8)2
8 12 4 x x x 9)2 4 3
2x x
x
10) 2 22 2 4 3 x x x x 11) 21 2 3 4 x x x x 12) 2 23 12 3 x x x x
13) 26 2 32 34 48 x x x x 14) 23 6 3 x x x x
15) 24 1 3 5 2 6 x x x x 16) 2 24 6 2 8 12 x x x x
17) 22 1 1 1 x x x x 18) 2 23 5 7 3 5 2 1 x x x x 19) 2 22 4 4 x x x
20) 2
2
3 4 92 3
3 3
xx
x
21) 2 23 4 9 x x x 22)
2
2
9 43 2
5 1
xx
x
23)6 3 3
4 4 2 x x x 24) 3 4 1 8 6 1 1 x x x x
25) 26 9 6 9 1 x x x x 26) 1 2 3 x x x 27)4 1 3
1 4 2x x
x x
28)1 1 1
1x
x x x x
* tm tp xc nh ca mi hm s sau:
1) 2 3 4 8 y x x x 2)2 1
2 1 2x x
yx x
3)
2 2
1 1
7 5 2 5y
x x x x
4) 2 5 14 3 y x x x 5)2
3 31
2 15
xy
x x
Cc dng ton c cha tham s:Bi1: Tm cc gi tr ca m mi biu thc sau lun dng:
a) 2 4 5 x x m b) 2 2 8 1 x m x m c) 22 4 2 x x m
d) 23 1 3 1 4m x m x m e) 21 2 1 3 2m x m x m
Bi 2: Tm cc gi tr ca m mi biu thc sau lun m:a) 24 1 2 1m x m x m b) 22 5 4m x x c) 2 12 5mx x
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d) 24 1 1 x m x m e) 2 22 2 2 1 x m x m f) 22 2 3 1m x m x m
Bi 3: Tm cc gi tr ca tham s m mi bt phng trnh sau nghim ng vi mi gi tr x:
a) 21 2 1 3 3 0m x m x m b) 2 24 5 2 1 2 0m m x m x
c)
2
2
8 200
2 1 9 4x x
mx m x m
d)
2
2
3 5 40
4 1 2 1x x
m x m x m
Bi 4: Tm cc gi tr ca m phng trnh:
a)
2
2 1 9 5 0 x m x m
c hai nghim m phn bitb) 22 2 3 0m x mx m c hai nghim dng phn bit.
c) 25 3 1 0m x mx m c hai nghim tri duBi 5: Tm cc gi tr ca m sao cho phng trnh :
4 2 21 2 1 0 x m x m
a) v nghim b) C hai nghim phn bit c) C bn nghim phn bitBi 6 : Tm cc gi tr ca m sao cho phng trnh: 4 2 21 1 0m x mx m c ba nghim phn bit
Bi 7: Cho phng trnh: 4 22 2 1 2 1 0m x m x m . Tm cc gi tr ca tham s m pt trn c:
a) Mt nghim b) Hai nghim phn bit c) C bn nghim
Bi 8: Xc nh cc gi tr ca tham s m mi bt phng trnh sau nghim ng vi mi x:a)
2
2
11
2 2 3
x mx
x x
b)
2
2
2 44 6
1
x mx
x x
c)
2
2
51 7
2 3 2
x x m
x x
Bi 9: Tm cc gi tr ca tham s m bt phng trnh sau v nghim:2 10 16 0
3 1
x x
mx m
Bi 10: Tm cc gi tr ca tham s m bt phng trnh sau c nghim:
a)
2 2 15 0
1 3
x x
m x
b)
2 3 4 0
1 2 0
x x
m x
PHNG TRNH, BPT V T
Bi 1. Gii cc pt sau:1) x 1 8 3x 1 22) x 2 4 2-xx
23) 3x 9 1 x-2x 24) 3x 9 1 x-2x
5) 3x 7- x 1 2 2 26) x 5 x 8 4 5x x Bi 2. Gii cc bpt sau:
21) x 12 7-xx 22) 21-4x-x x 3 23) 1-x 2x 3 5 0x 24) x 3 10 x-2x
25) 3 -x 6 2(2x-1) 0x 26) 3x 13 4 2-x 0x
7) x 3- 7-x 2x-8 8) 2x 3 x 2 1 29) 2x x 1 x 1 10) 2-x 7-x - -3-2x
11) 11-x - x-1 2 4
12) - 2-x 22-x
2x 16 513) x-3
3 x-3x
14) 1-4x 2x 1
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69) 4 2 4 2 2x +x 1 + x x +1 2x 70) x+3 x1< x2 71) x+1 x1 x 72) 5x+1 4x1 3 x 73) x+1 > 3 x+4 74) x+2 3x< 52x
75) 2 2 2x +x+1+ x x+1 2x +6x+2 76) 6x + 1 2x + 3 < 8x 4x + 2
77) x + x + 9 x + 1 + x + 4 78) 3 312 x + 14 + x 2
79)3 3
4 x + x + 8 2 80)2
x 1 x < 0 81)
2
2
9x 40
5x 1 82) 22x 5 2x 5x + 2 0
83) 2 2(x 4x + 3) x 4 >0 84)2
(x1)x(x + 2) 0
(x2)
85) 2 2(x 3x) 2x 3x 2 0 86) 2 2( 2) x + 4 x 4x
87)2
2
3(4x 9)2x+3
3x 3 88) 2 2(x 3) x + 4 x 9
89)2
29x 4 3x+25x 1
90) 2 2x(x 4) 4x x 4 (2 x)
91)2x
3x 2 1 x3x 2
92)2
2 2
2
xx x 4+ 4x
2 4 x
93) x+34x+1 3x25
94) 2
2 2
2
x3x 2x +1 25 x
5 + 25 x
95) 22
40x + x +16
x +16 96) 2 23x +5x+7 3x +5x+2 >1
97)2
24x < 2x + 9(1 1 + 2x )98) 2x > 2x + 22x + 1 1
99) 2 24(x + 1) < (2x + 10)(1 3 + 2x) 100)2
2
2xx + 21
(3 9 + 2x )
101)2
2
x> x 4
(1 + 1 + x )102) 2 29(x + 1) (3x + 7)(1 3x + 4)
103) (x1) 2x 1 3(x1) 104)2x
> 2x + 22x + 1 1
105)
2
2
x
> x 4(1 + 1 + x ) 106)
2
2
4x
< 2x + 9(1 1 + 2x )
107)2
2
2xx + 21
(3 9 + 2x ) 108) 2 24(x + 1) < (2x + 10)(1 3 + 2x)
109) 2 2x + 4x (x + 4) x 2x + 4 110) 2 29(x + 1) (3x + 7)(1 3x + 4)
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PHNG PHP I BIN S:Bi 1. Gii cc pt sau:
2 21) 3x 5 8- 3x 5 1 1x x 2 22) x 9- x 7 2
21 x 21 213)
x21 x 21
x
x
2 2 24) 3x 6 16 2 2 x 2 4 x x x x
5) (x 5)(x-2) 3 x x 3 0 6) 2(x 1)(x 4) 5 x 5 28x
7) 2 23x 5 7- 3x 5 2 1x x 8) 2(x 4)(x 1)-3 x 5 2 6x 9) 24 4 x 2 x x 2x 8 10) 2 22x x 5x 6 10x 15
11) 2 22x 4x 3 3 2x x 1 12) 26 (x 2)(x 32) x 34x+48
13) 2x(x + 3) 6 x 3x 14) 2(x +4)(x +1) 3 x +5x+2 x x+1
17) 2(x +1)(x +4) < 5 x +5x+28 18) 2 2x + 2x + 5 4 2x + 4x+3
19) 2 22x + x 5x 6 >10x+15 20) 4x x1 3 >
x1 4x 2
21) x x+1 2 >3x+1 x
22) 46x 12x 12x 2. 0x2 x2 x2
23) 3 6x2 x2 x2+2. + 4 0x+1 x+1 x+1
24)5 1
5 x+ < 2x+ +42x2 x
25) 2 14 x+ < 2x+ +22xx
26) 3 13 x+ < 2x+ 72x2 x
27) 3x > 1 + x1 28) 3 2(x + 1) + (x + 1) + 3x x+1 > 0
29) x 1 + x + 3 + 2 (x 1)(x + 3) > 4 2x 30) 2 2x + 1 x x. 1 x
31) x + 5 + x 3 < 1 + (x + 5)(x 3) 32) 2x 35x+ > 12x 1
33) 27x+7 + 7x6 + 2 49x +7x 42 3 5
x 4
35) 22x + x + x + 7 + 2 x + 7x 35 36)2 2
1 3x+1>
1x 1x
37) 2 2 2x 4x + 6 + x 4x + 8 2x 8x + 32 38)2
2 2
2 2
5a2(x+ x +a )
x +a
39) 2 2x 1 2x x +2x 40) 2 2x 1 2x x 2x
41) 2x1 x( x1 x ) + x x 42)2 2
1 3x+ 1 >
1 x 1 x
43) 3 3(4x 1) x +1 2x + 2x + 1 44) 22x +12x +6 2x 1 > x +2
45) x 1 + x + 3 + 2 (x 1)(x + 3) > 4 2x 46) 22x 6x + 8 x x 2
47) x + 5 + x 3 < 1 + (x + 5)(x 3) 48) 3 2 2x 2x + x x x + x 2x
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8/9/2019 Phu Dao Dai So 10 CA Nam
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Ebook4Me.Net
49) 27x+7 + 7x6 +2 49x +7x42 3 5
x 458)
22 2
2 2
5a2(x + x + a ) (a 0)
x + a
PHNG PHP HM S:1) x+1 + 2x+3 > 5 2) x+9 + 2x+4 > 5
3) 2x+1 > 7 x 4) 31 x < x + 5
5) 2 3x + 1 1 2x + x x 6) 2 2x 2x + 3 x 6x + 11 > 3 x x 1
7) 2x + x 1 1 8) 2x 1 + x 1 (x + 1)(3 x)
9) 2 2 2 23x 7x + 3 + x 3x + 4 > x 2 + 3x 5x 1 PHNG PHP NH GI:
(nh gi bng BT):
1) 2 2x + x 1 + x x +1 x+1 2)2x
1 + x + 1 x 2 +4
3
3) 2 2x x 1+ x + x 1 2 4) 1 + x 1 x x
5) 2 22x + 4 + 2 2 x 2 6 6) 22x 10x + 16 x 1 x 3
7) 2 4 22 x x + x + 1 x 1 + 2 8) x + 2 x 1 + x 2 x 1 2
9) 2 2 4 3 2(2x 3x + 1) 4x 20x + 25x < 2x + 1 10) 3 2 3x 2 2 x 11) x x 2 >
1 x + x 1 x x x
(nh gi bng o hm):
1) 5 5(1 x) + (1 +x) 4 2 2)2x
1 + x + 1 x 2 4
3) 3 20023x +1 + 2x +4 < 3 x189
4) 3 22x + 3x + 6x + 16 > 2 3 + 4 x
5)2 2 3 23
x + (1 x ) 27