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TTTRRRNNNGGG TTTHHHPPPTTT CCCHHHUUU VVVNNN AAANNNTTT TTTOOONNN
GGGVVV::: DDDnnng gg PPPhhhccc SSSaaannnggg
On tap Tot nghiep
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Dng Phc Sang - 1 - THPT Chu Vn An
PPPhhhnnn III... KKKHHHOOO SSST TT HHHM MM SSS VVV BBBIII TTTOOONNN LLLIIINNN QQQUUUAAANNN
1. Hm s bc ba, hm s trng phng v cc vn lin quana) Kho st s bin thin v v th hm s
1 Tp xc nh: D = 2 Tnh y
3 Cho 0y = tm cc nghim0
x (nu c).
4 Tnh hai gii hn: lim ; limx x
y y +
5 V bng bin thin ca hm s.6 Nu s ng bin, nghch bin v cc tr (nu c) ca hm s.7 Tm im un (i vi hm s bc ba).8 Lp bng gi tr.9 V th hm s v nu nhn xt.
3 2 ( 0)y ax bx cx d a = + + +
S nghim ca phngtrnh 0y = 0a > 0a <
0y = c 2 nghimphn bit
0y = c nghim kp
0y = v nghim
th hm s bc ba lun i xng qua im un
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Ti liu tham kho - 2 - n tp tt nghip mn Ton
4 2 ( 0)y ax bx c a = + +
S nghim ca phngtrnh 0y = 0a > 0a <
0y = c 3 nghimphn bit
0y = c 1 nghim duy
nht
th hm s trng phng lun i xng qua trc tungb) Vit phng trnh tip tuyn (dng 1 bit to tip im M0)
1 Ch r0
x v0
y (honh & tung ca im M0)
2 Tnh0
( )f x
3
Cng thc: 0 0 0( )( )y y f x x x
= c) Vit phng trnh tip tuyn (dng 2 bit trc h s gc k)
1 Lp lun c c0
( ) f x k = (*)
2 Thay0
( )y x vo (*) tm0
x
3 C0
x , tm0
y v dng cng thc
0 0 0( )( )y y f x x x =
Lu : Tip tuyn song song vi y ax b= + c h s gc k= a Tip tuyn vung gc vi ( 0)y ax b a = + c h s
gc 1a
k =
d) Bin lun s nghim ca phng trnh bng th (C):y=f(x)1 a phng trnh v dng: ( ) ( ) f x BT m = 2 Lp lun: s nghim ca phng trnh cho bng vi s giao
im ca th ( ) : ( )C y f x = v ng thng : ( )d y BT m = .3 V 2 ng ln cng 1 h trc to v lp bng kt qu
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Dng Phc Sang - 3 - THPT Chu Vn An
Lu : nu bi ton ch yu cu tm cc gi tr ca m phng
trnh c ng 3 nghim, 4 nghim, ta khng cn lp bng ktqu nh trn m ch cn ch r cc trng hp tho .e) S tng giao gia th (C):y=f(x) v ng thng d: y= ax+ b
1 Lp phng trnh honh giao im ca ( )C v d:
( ) f x ax b= + (*)2 Lp lun: s giao im ca ( )C v dbng vi s nghim ca (*)
3m s nghim ca (*) suy ra s giao im ca ( )C v d
V D MINH HOBi 1: Cho hm s 3 26 9 1y x x x = + +
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca ( )C ti giao im ca ( )C vi
trc tung.c) Tm cc gi tr ca tham sm phng trnh sau y c
nghim duy nht: 3 26 9 0x x x m + + =
Bi giiCu a: Hm s 3 26 9 1y x x x = + + Tp xc nh: D= R
o hm: 23 12 9y x x = +
Cho 20 3 12 9 0 1y x x x = + = = hoc 3x =
Gii hn: lim ; limx x
y y +
= = +
Hm s ng bin trn cc khong (;1) v (3;+)Hm s nghch bin trn khong (1;3) th hm s c im cc i (1;5)D , im cc tiu (3;1)T
Cho6 12. 0 2 3y x y x y = = = = . im un (2;3)I
Bng bin thin:(ch : do a> 0) x 1 3 + y + 0 0 +
y5 +
1
m BT(m) S giao im S nghim pt . .
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Ti liu tham kho - 4 - n tp tt nghip mn Ton
Bng gi tr: x 0 1 2 3 4y 1 5 3 1 5
th hm s l mt ng cong i xngqua im (2;3)I nh hnh v bn y:
Cu b: Cho 0 (0) 1x y= = .Giao im ca ( )C vi trc tung l: (0;1)A
(0) 9f = Phng trnh tip tuyn ca ( )C ti A l:
1 9( 0) 9 1y x y x = = +
Cu c: Ta c, 3 2 3 26 9 0 6 9x x x m x x x m + + = + = 3 2
6 9 1 1x x x m + + = (*) Phng trnh (*) c nghim duy nht khi v ch khi th ( )C vng thng : 1d y m= ct nhau ti 1 im duy nht
1 5 4
1 1 0
m m
m m
> < < >
Bi 2: Cho hm s 2 33 2y x x= a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca ( )C ti cc giao im ca ( )C
vi trc honh.
c) Bin lun theo as nghim phng trnh: 3 24 6 3 0x x a = Bi gii
Cu a: Hm s 2 33 2y x x= Tp xc nh: D =
o hm: 26 6y x x =
Cho 20 6 6 0 0y x x x = = = hoc 1x =
Gii hn: lim ; limx x
y y +
= + =
Hm s ng bin trn khong (0;1)
Bng bin thin:(ch : do a< 0)
x 0 1 + y 0 + 0
y+ 1
0
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Dng Phc Sang - 5 - THPT Chu Vn An
Hm s nghch bin trn cc khong ( ; 0) v (1; )+ th hm s c im cc i (1;1)D , im cc tiu (0; 0)O
Cho 1 12 2
6 12 . 0y x y x y = = = = . im un 1 12 2
( ; )I
Bng gi tr: x 12 0 12 1 12 y 1 0 1
21 0
th hm s l mt ng cong i xngqua im 1 1
2 2( ; )I nh hnh v bn y:
Cu b: Cho 2 30 3 2 0y x x= = 3
2
0x
x
= =
Giao im ca ( )C vi trc honh l: (0;0)O v 32
( ;0)B
Ti (0;0)O : (0) 0f = , phng trnh tip tuyn l: 0y =
Ti 32
( ;0)B : 3 92 2
( )f = , phng trnh tip tuyn l:
279 3 92 2 2 4
0 ( )y x y x = = +
Cu c: Ta c,3 2 2 3 2 34 6 3 0 6 4 3 3 2x x a x x a x x = = 3
2a= (*)
S nghim phng trnh (*) bng vi s giao im ca th ( )C
v ng thng 32
:d y a= , do ta c bng kt qu sau y:
a 32
a S giao imca ( )C v d
S nghim caphng trnh (*)
2
3a < 3
2 1a > 1 123
a = 32
1a = 2 2
23
0a < < 32
0 1a< < 3 3
0a = 32 0a = 2 2
0a > 32
0a < 1 1
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Ti liu tham kho - 6 - n tp tt nghip mn Ton
Bi 3:a) Kho st v v th ( )C ca hm s 3 23 32
x x xy
+ +=
b) Vit phng trnh tip tuyn vi th ( )C bit tip tuyn song
song vi ng thng
3
2:y x
= c) Tm to cc giao im ca ( )C vi ng thng
23y 2x= +
Bi gii
Cu a:3 23 3
2
x x xy
+ += Tp xc nh: D =
o hm23 6 3
0,2
x xy x
+ + = do hm s lun ng
bin trn v khng t cc tr. Gii hn: lim ; lim
x xy y
+= = +
Bng bin thin:
12
3 3 0 1y x x y = + = = =
im un 12
( 1; )I
Bng gi tr: x 3 2 1 0 1y 9
2 1 1
20 7
2
th hm s l ng cong i xng qua im12( 1; )I
Cu b: Tip tuyn ca ( )C song song vi ng thng 32
: y x = c h
s gc 30 2( )k f x= =
20 0
3 6 3
2
x x+ + = 3
22 00 0
0
03 6 0
2
xx x
x
= + = =
Vi0
0x = th0
(0) 0y y= = , tip tuyn tng ng l3 32 2
0 ( 0)y x y x = = (trng vi )
x 1 + y + 0 +
y
+
12
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Dng Phc Sang - 7 - THPT Chu Vn An
Vi0
2x = th 0 ( 2) 1y y= = , tip tuyn tng ng l
3 32 2
1 ( 2) 2y x y x + = + = + (song song vi )
Vy, tip tuyn tho l2
3y 2x= +
Cu c: Honh giao im (nu c) ca ( )C v23y 2x= + l nghim
phng trnh3 23 3
2
x x x+ += 3 2
32 3 3 3 4
2x x x x x + + + = +
3 2 21
3 4 0 ( 1)( 4 4) 02
xx x x x x
x
= + = + + = =
721x y= = v 2 1x y= =
Vy, ( )C v 32
: 2d y x= + ct nhau ti 2 im:
( )721;A v ( 2; 1)B
Bi 4:a) Kho st v v th ( )C ca hm s: 4 22 3y x x= b) Vit phng trnh tip tuyn vi th ( )C ti im trn ( )C
c honh xl nghim ca phng trnh ( ) 20f x = c) Tm cc gi tr ca tham sm phng trnh sau y c nhiu
hn hai nghim: 4 22 0x x m + = Bi gii
Cu a:Hm s 4 22 3y x x= Tp xc nh: D =
34 4y x x = Cho 30 4 4 0 0; 1y x x x x = = = =
Gii hn: lim ; limx x
y y +
= + = +
Bng bin thin:
x 1 0 1 +
y 0 + 0 0 +y + 3 +
4 4
Hm s ng bin trn cc khong trn (1;0), (1;+) v nghch
bin trn cc khong (;1), (0;1).
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Ti liu tham kho - 8 - n tp tt nghip mn Ton
th hm s c im cc i (0; 3)D
v hai im cc tiu1 2( 1; 4), (1; 4)T T
Bng gi tr:
x 2 1 0 1 2 y 3 4 3 4 3 th hm s l ng cong i xngqua trc tung nh hnh v
Cu b:Ta c, 2 2 212 4 20 12 24 2 2y x x x x = = = = =
p s: 4 2 11y x= v 4 2 11y x= (hc sinh t gii)
Cu c:Ta c, 4 2 4 22 0 2 3 3x x m x x m + = = (*) Phng trnh (*) c nhiu hn 2 nghim khi v ch khi ( )C v
: 3d y m= ct nhau ti nhiu hn 2 im (3 hoc 4 im)
3 3 00 1
3 4 1
m mm
m m
4 m3 > 1 0 0m= 4 m3 = 1 2 2
0 < m< 4 3 < m3 < 1 4 4m= 0 m3 = 3 3 3m< 0 m3 < 3 2 2
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Dng Phc Sang - 9 - THPT Chu Vn An
BI TP V HM S BC BA V HM S TRNG PHNGBi 6: Cho hm s 3 3 1y x x= + c th l ( )C
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti im thuc ( )C c honh bng 2.c) Vit pttt vi ( )C bit tip tuyn c h s gc bng 9.d) Tm iu kin ca m phng trnh sau c 3 nghim phn bit:
3 3 1 2 0x x m+ + = .
Bi 7: Cho hm s 3 21 32 2
2y x x= +
a) Kho st s bin thin v v th ( )C ca hm s.
b) Vit pttt vi ( )C song song vi ng thng d: 92
2y x= +
c) Tm cc gi tr ca k phng trnh sau y c nghim duy
nht: 3 23 4 0x x k = Bi 8: Cho hm s 3 22 3 1y x x= +
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti giao im ca ( )C vi trc honh.c) Vit pttt vi ( )C bit tip tuyn song song vi : 12 1d y x=
d) Bin lun theo ms nghim phng trnh: 3 22 3 2 0x x m+ + =
Bi 9: Cho hm s 3 21 3 53 2 2
y x x= + c th l ( )C
a) Kho st s bin thin v v th ( )C ca hm s.
b) Vit pttt vi ( )C ti im trn ( )C c honh xtho 1y = c) Tnh din tch hnh phng gii hn bi ( )C v : 2 0d y = .d) Tm cc gi tr ca m phng trnh sau c nghim duy nht
3 22 9 6 0x xe e m + =
Bi 10: Cho hm s 3 213
y x x=
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt ca ( )C ti im trn ( )C c tung bng 0.c) Vit pttt ca ( )C song song vi ng thng 8 3y x= d) Tm cc gi tr ca a phng trnh sau y c nghim duy
nht: 3 23 log 0x x a =
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Ti liu tham kho - 10 - n tp tt nghip mn Ton
Bi 11: Cho hm s 3 22 3 1y x x= (*)a) Kho st s bin thin v v th ( )C ca hm s.b) Tm to giao im ca ( )C vi ng thng d: 1y x=
c) Bin lun theo ms nghim ca phng trnh3 24 6 1 0x x m + =
Bi 12: Cho hm s 3 23 2y x x= + , m l tham s.a) Kho st s bin thin v v th ( )C ca hm s.
b) Vit pttt ca ( )C vung gc vi ng thng d: 1 13 3
y x=
c) Tm cc gi tr ca a ng thng 2y ax= + ct ( )C ti ba
im phn bit.Bi 13: Cho hm s 3 23 2y x x= + c th ( )C a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca ( )C ti im A(0; 2)c) Vit pttt ca ( )C bit tip tuyn song song vi 9 4 4 0x y = d) Bin lun theo ms giao im ca ( )C v : 2d y mx =
Bi 14: Cho hm s 34 3 1y x x= , c th l ( )C a) Kho st s bin thin v v th ( )C ca hm s.
b) Tm m phng trnh 34 3 1x x m = c ng 3 nghim.c) Vit pttt vi ( )C ti giao im ca ( )C vi trc honh.
d) Vit pttt vi ( )C bit tip tuyn vung gc vi 172
:d y x=
Bi 15: Cho hm s 3 22 6 6 2y x x x = + a) Kho st s bin thin v v th ( )C ca hm s.
b) Tnh din tch hnh phng gii hn bi ( )C , Ox, 1, 2x x= =
Bi 16: Cho hm s 2 2(2 )y x x= a) Kho st s bin thin v v th ( )C ca hm s.
b) Vit pttt vi ( )C ti im trn ( )C c honh bng 2 c) Vit pttt vi ( )C bit tip tuyn c h s gc bng 24.d) Tm cc gi tr ca tham sm phng trnh sau c 4 nghim
4 22 0x x m + =
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Dng Phc Sang - 11 - THPT Chu Vn An
Bi 17: Cho hm s 4 22 3y x x= + a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt ca ( )C ti im trn ( )C c tung bng 5.
c) Tm iu kin ca m phng trnh sau y c ng 2 nghim:4 22 3 2 0x x m+ + + =
Bi 18: Cho hm s 12
y = 4 23x x + 32
c th ( )C .
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C bit tip tuyn c h s gc bng 8.
c) Tm m phng trnh sau c 4 nghim: 4 26 log 0x x m + =
Bi 19: Cho hm s 2 2(1 ) 6y x= c th ( )C a) Kho st s bin thin v v th ( )C ca hm s.
b) Bin lun theo ms nghim ca phng trnh : 4 22x x m = c) Vit pttt ca ( )C bit tip tuyn vung gc vi 1
24:d y x=
Bi 20: Cho hm s 14
y = 4 22 1x x+
a) Kho st s bin thin v v th ( )C ca hm s.
b) Tm m phng trnh 4 28 4x x m + = c nhiu hn 2 nghimc) Vit phng trnh tip tuyn ca th ( )C ti im trn ( )C
c honh l nghim ca phng trnh ( ) 10y x =
Bi 21: Cho hm s 14
y = 4 22x x
a) Kho st s bin thin v v th ( )C ca hm s.
b) Vit pttt ca ( )C song song vi1: 15 2012d y x= + .
c) Vit pttt ca ( )C vung gc vi2
:d 845
2012y x= +
d) Tm m phng trnh 4 28x x m + = c 4 nghim phn bit.
Bi 22: Cho hm s 4 2 ( 1)y x mx m = + c th ( )Cm a) Tm m th hm s i qua im ( 1; 4)M b) Kho st v v th ( )C ca hm s khi 2m = .c) Gi ( )H l hnh phng gii hn bi ( )C v trc honh. Tnh th
tch vt th trn xoay to ra khi quay ( )H quanh trc honh.
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Ti liu tham kho - 12 - n tp tt nghip mn Ton
2. Hm s nht bin v cc vn lin quana) Kho st s bin thin v v th hm s( 0, 0)c ad cb
ax by
cx d
+=
+
1 Tp xc nh: { }\ dc
D =
2 Tnh2( )
ad cby
cx d
=+
v khng nh y dng hay m, dc
x
3Suy ra hm s ng bin hay nghch bin trn mi khong xc
nh ( ; ),( ; )d dc c
+ v khng t cc tr.
4 Tnh cc gii hn v tm hai tim cn:Tnh lim
xy
a
c= v lim
xy
a
+ c= , suy ra
ay
c= l TCN
Tnh( )
limd
cx
y
v( )
limd
cx
y+
, suy rad
xc
= l TC
5 V bng bin thin ca hm s.6 Lp bng gi tr.
7V th hm s (c 2 tim cn) v nu nhn xt.
( 0, 0)ax b
y c ad cb cx d
+=
+
0y > 0y <
th hm s nht bin gm hai nhnh ring bitlun i xng nhau qua giao im ca hai ng tim cn
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Dng Phc Sang - 13 - THPT Chu Vn An
b) Vit phng trnh tip tuyn (dng 1 bit to tip im M0)
1 Ch r0
x v0
y (honh & tung ca im M0)
2 Tnh0
( )f x
3 Cng thc: 0 0 0( )( )y y f x x x = c) Vit phng trnh tip tuyn (dng 2 bit trc h s gc k)
1 Lp lun c c0
( ) f x k = (*)
2 Thay0
( )y x vo (*) tm0
x
3 C0
x , tm0
y v dng cng thc0 0 0
( )( )y y f x x x =
Lu : Tip tuyn song song vi y ax b= + c h s gc k= a Tip tuyn vung gc vi ( 0)y ax b a = + c h s
gc 1a
k =
d) S tng giao gia th (C):y=f(x) v ng thng d: y= ax+ b1 Lp phng trnh honh giao im ca ( )C v d:
( ) f x ax b= + (*)2 Lp lun: s giao im ca ( )C v dbng vi s nghim ca (*)
3 m s nghim ca (*) suy ra s giao im ca ( )C v dV D MINH HO
Bi 23: Cho hm s 2 11
xy
x
+=
+
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca ( )C ti im trn ( )C c tung
bng 5
2
c) Chng minh rng ng thng : 2d y x m = + lun ct th( )C ti 2 im phn bit.
Bi gii
Cu a: Hm s2 1
1
xy
x
+=
+ Tp xc nh: \{ 1}D =
o hm:2
10, 1
( 1)
y x
x
= >
+
, do hm s ng bin
trn cc khong ( ; 1) , ( 1; ) + v khng t cc tr.
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Ti liu tham kho - 14 - n tp tt nghip mn Ton
Gii hn v tim cn:
lim 2 ; lim 2x x
y y +
= = y= 2 l tim cn ngang.
( 1) ( 1)
lim ; limx x
y y +
= + = 1x = l tim cn ng.
Bng bin thin:x 1 +
y + +
y+
22
Bng gi tr:
x 2 32
1 12
0
y 3 4 0 1 th hm s gm hai nhnh i xngnhau qua im ( 1;2)I nh hnh v
Cu b: Vi 52
y = th2 1 5
2(2 1) 5( 1) 31 2
xx x x
x
+= + = + =
+
Ta c2
1 14( 2)
( 3)f
= =
Vy, tip tuyn ca ( )C ti 52( 3; )M l:5 1 1 132 4 4 4
( 3)y x y x = + = +
Cu c:Honh giao im (nu c) ca ( )C v dl nghim phng trnh2 1
2 2 1 ( 2 )( 1)1
xx m x x m x
x
+= + + = + +
+, 1x
22 (4 ) 1 0x m x m + + = (*) ( 1x = khng tho (*))
Bit thc ca phng trnh (*):2 24 12 ( 2) 8 0,m m m m = + = + >
Do 0 > nn (*) lun c 2 nghim phn bit, t ( )C v dlun c 2 im chung phn bit.
Bi 24:a) Kho st v v th ( )C ca hm s 32
xy
x
=
b) Vit pttt ca ( )C bit tip tuyn song song vi :d y x=
c) Tm cc gi tr ca m ng thng :d y x m = + ct th( )C ti 2 im phn bit.
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Dng Phc Sang - 15 - THPT Chu Vn An
Cu a: Hm s3 3
2 2
x xy
x x
= =
+ Tp xc nh: \ {2}D =
o hm:2
10, 2
(2 )y x
x
= <
, do hm s nghch bin
trn cc khong ( ;2) , (2; )+ v khng t cc tr. Gii hn v tim cn:
lim 1 ; lim 1x x
y y +
= = 1y = l tim cn ngang.
2 2
lim ; limx x
y y +
= = + 2x = l tim cn ng.
Bng bin thin:x 2 +
y
y1
+
1 Bng gi tr:
x 0 1 2 3 4y 3
22 0 1
2
th hm s gm hai nhnh i xng
nhau qua im (2; 1)I nh hnh v
Cu b: V tip tuyn song song vi ng thng y x= nn c h s
gc0
( ) 1k f x= =
20
11
(2 )x
=
2
0(2 ) 1x = 0 0
0 0
2 1 1
2 1 3
x x
x x
= = = =
p s:c 2 tip tuyn tho l 1y x= v 3y x= + Cu c: Phng trnh honh giao im ca ( )C v d:
3
2
xx m
x
= +
2 ( 3) 2 3 0x m x m + + + = (*)
( )C v d ct nhau ti 2 im phn bit khi v ch khi phng
trnh (*) c 2 nghim phn bit 20 2 3 0m m > > ( ; 1) (3; )m +
Vy vi ( ; 1) (3; )m + th th ( )C v ng thng:d y x m = + ct nhau ti 2 im phn bit.
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Ti liu tham kho - 16 - n tp tt nghip mn Ton
BI TP V HM S NHT BIN
Bi 25: Cho hm s 2 11
xy
x
+=
a) Kho st s bin thin v v th ( )C ca hm s.
b) Vit pttt vi ( )C bit tip tuyn c h s gc bng 3.c) Vit pttt vi ( )C ti im trn ( )C c tung bng 7
2
d) Tm m : ( 1) 2d y m x = + + ct ( )C ti 2 im phn bit.
Bi 26: Cho hm s 2 11
xy
x
+=
+
a) Kho st s bin thin v v th ( )H ca hm s.
b) Lp phng trnh tip tuyn ca ( )H bit tip tuyn song songvi ng phn gic ca gc phn t th nht.c) Vit pttt vi ( )H ti im trn ( )H c honh bng 3 .d) Tm m ng thng 1y mx= + ct ( )C ti 2 im phn bit.
Bi 27: Cho hm s 2 12
xy
x
=
a) Kho st s bin thin v v th ( )C ca hm s.
b) Vit pttt vi ( )C bit tip tuyn c h s gc bng34
c) Chng minh rng vi mi gi tr ca tham s m ng thngy x m= lun ct th ( )C ti hai im phn bit.
Bi 28: Cho hm s 32y1x
= +
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi th ( )C ti giao im ca ( )C vi trc honh.
c) Tm m ng thng :d y m x = ct ( )C ti 2 im phn bit
Bi 29: Cho hm s 23
xy
x
+=
c th ( )C .
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti im trn ( )C c honh bng 1.c) Vit pttt vi ( )C ti im trn ( )C c tung bng 3
2
d)Vit pttt vi ( )C bit tip tuyn c h s gc bng 5
4
e) Xc nh to giao im ca ( )C v 3 2y x= +
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Dng Phc Sang - 17 - THPT Chu Vn An
Bi 30: Cho hm s 21
yx
=+
c th l ( )C .
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit phng trnh tip tuyn ca th ( )C ti cc giao im
ca ( )C vi ng thng : 2 1d y x= c) Tm gi tr ln nht ca hm s trn on [0;2]
d) Vit pttt ca ( )C bit tip tuyn song song vi 1 32 2
y x= +
e) Tnh din tch hnh phng gii hn bi ( )C trc honh v haing thng x= 0, x= 2.
Bi 31: Cho hm s 11
xy
x
=
+
c th ( )C .
a) Kho st s bin thin v v th hm s.b) Tm im Mtrn trc honh m tip tuyn ca ( )C i qua im
Msong song vi ng thng d: y= 2x
Bi 32: Cho hm s 21
xy
x
=
+
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti giao im ca ( )C vi : 2 3d y x= .
c) Vit pttt ca ( )C vung gc vi ng thng21y 2012x= +
d) Tm m ng thng d: 2y mx= + ct c hai nhnh ca ( )C .
Bi 33: Cho hm s 2 31
xy
x
=
a) Kho st s bin thin v v th ( )C ca hm s.b) Tnh din tch hnh phng gii hn bi ( )C , Ox v 2x = .
c) Vit phng trnh cc ng thng song song vi ng thng3y x= + ng thi tip xc vi th ( )C
Bi 34: Cho hm s 3 41
xy
x
+=
a) Kho st s bin thin v v th ( )C ca hm s.b) Vit pttt vi ( )C ti giao im ca ( )C vi trc tung.c) Vit pttt vi ( )C ti cc giao im ca ( )C vi : 2 4d y x=
d) Tm a ng thng : 3y ax = + th ( )C khng giao nhaue) Tm tt c cc im trn ( )C c to u l cc s nguyn.
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Ti liu tham kho - 18 - n tp tt nghip mn Ton
3. Tm GTLN, GTNN ca hm s y= f(x) trn on [a;b]1Hm s ( )y f x= lin tc trn on [a;b].
2 Tnh ( )y f x = .
3
Cho 0y = tm cc nghim [ ; ]ix a b (nu c) v cc s[ ; ]
jx a b lm cho y khng xc nh (nh loi cc s [ ; ]x a b
l)
4 Tnh cc gi tr ( )i
f x , ( )j
f x v ( ), ( )f a f b
(khng c tnhf ca cc xl
b loi)
5Chn kt qu ln nht v kt qu nh nht t bc 4 kt lunv gi tr ln nht v gi tr nh nht ca hm s trn on [a;b].
4. iu kin hm s c cc tr(tm tt)
Nu 0
0
( ) 0
( ) 0
f x
f x
= th hm s ( )y f x= t cc tiu ti
0x
Hm s 3 2y ax bx cx d = + + + c cc i, cc tiu 0y >
Hm s 4 2y ax bx c = + + c cc i, cc tiu . 0a b < 5. iu kin hm s n iu trn tng khong xc nh
Hm s 3 2y ax bx cx d = + + + ng bin trn
00,
0yy x
a
>
Hm s 3 2y ax bx cx d = + + + nghch bin trn
00,
0yy x
a
> (khng c du =)
Hm sax b
y
cx d
+=
+
nghch bin trn tng khong xc nh
0, 0y x D ad cb < < (khng c du =)
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Dng Phc Sang - 19 - THPT Chu Vn An
V D MINH HOBi 35: Tm gi tr ln nht v gi nh nh nht ca hm s:
a) 3 28 16 9y x x x = + trn on [1;3]
b) 2 4 ln(1 )y x x= trn on [3;0]c) 3 22 ln 3 ln 2y x x= trn on 2[1; ]e
d) 2( 1)xy e x x = trn on [0;2]
Bi giiCu a: Hm s 3 28 16 9y x x x = + lin tc trn on [1;3]
o hm: 23 16 16y x x = +
Cho2
0 3 16 16 0y x x = + =
loai
nhan43
4 [1; 3] ( )
[1; 3] ( )
x
x
= =
Trn on [1;3] ta c: ( ) ; ;43 27 (1) 0 (3) 613 f f f = = =
Do 1327
6 0 < < nn[1;3]
min (3) 6x
y f
= = v[1;3]
maxx
y
( )4 133 27f= = Cu b: Hm s 2 4 ln(1 )y x x= lin tc trn on [3;0]
24 2 2 4
2 1 1
x x
y x x x
+ +
= + =
Cho(nhan)
(loai)2
1 [ 3;0]0 2 2 4 0
2 [ 3;0]
xy x x
x
= = + + = =
Trn on [2;0]: ; ;( 1) 1 4 ln 2 ( 3) 9 8 ln 2 (0) 0 f f f = = =
Do16
1 4 ln 2 ln 0e = < v2
9 8 ln 2 1 8 ln 0e = + > nn
[ 3;0]
min ( 1) 1 4 ln 2x
y f
= = v[ 3;0]
max ( 3) 9 8 ln 2x
y f
= =
Cu c:Hm s 3 22 ln 3 ln 2y x x= lin tc trn on 2[1; ]e t lnt x= th 2[1; ] [0;2]x e t , hm s tr thnh
3 2( ) 2 3 2y g t t t = = c 20 [0;2]
( ) 6 6 01 [0;2]
tg t t t
t
= = = =
Trn on [0;2]: (0) 2 ; (1) 3 ; (2) 2g g g= = =
Do 3 2 2 < < nn 2[1; ]min (1) 3x e y g = = v 2[1; ]max (2) 2x e y g = =
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Ti liu tham kho - 20 - n tp tt nghip mn Ton
Cu d: p s:[0;2]min (1)y f e= = v 2
[0;2]max (2)y f e= =
Bi 36: Tm iu kin ca tham sm hm s 3 2 4 3y x mx x = + + + a) ng bin trn b) C cc i v cc tiu
Bi giiCu a: 3 2 4 3y x mx x = + + + (*)
Tp xc nh: D= R
o hm: 23 2 4y x mx = + + c 2 12y
m =
Hm s (*) ng bin trn 0,y x
2
3 00 2 30 12 0y
am
m
>>
Vy, vi 2 3 ; 2 3m th hm s (*) ng bin trn
Cu b:Hm s (*) c cc i v cc tiu 0y = c 2 nghim phn
bit 20 12 0 ( ; 2 3) (2 3; )y
m m > > +
Vy vi ( ; 2 3) (2 3; )m + th hm s (*) c cc i vcc tiu.
Bi 37: Tm iu kin ca m hm s 3 2 23 ( 1) 2y x mx m x = + + t cc i ti
02x =
Bi gii
Cu a: 3 2 23 ( 1) 2y x mx m x = + + (*)
Tp xc nh: D=R
o hm: 2 2( ) 3 6 ( 1)y f x x mx m = = +
( ) 6 6y f x x m = =
Hm s (*) t cc i ti0
2x = khi v ch khi2(2) 0 {1;11}12 11 0
11(2) 0 212 6 0
f mm mm
f mm
= + = = < >
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Dng Phc Sang - 21 - THPT Chu Vn An
Bi 38: Chng minh rng nu sinx
xy
e= th 2 2 0y y y + + =
Bi gii
Hm s
sin
.sin
x
x
x
y e xe
= = c tp xc nh D = ( ) . sin .(sin ) (cos sin )x x xy e x e x e x x = + =
( ) (cos sin ) (cos sin ) 2 cosx x xy e x x e x x e x = + =
2 2 2 cos 2 (cos sin ) 2 sin 0x x xy y y e x e x x e x + + = + + =
Vy, vi .sinxy e x= th 2 2 0y y y + + =
BI TP V CC VN KHC LIN QUAN HM SBi 39: Tm gi tr ln nht, gi tr nh nht ca cc hm s sau y
a) 3 2( ) 2 3 12 10 f x x x x = + trn on [ 2; 0]
b) 5 4 3( ) 5 5 1 f x x x x = + + trn on [1;2]
c) 4 3 2( ) 2 1 f x x x x = + trn on [1;1]
d) 5 3( ) 5 10 1 f x x x x = + trn on [2;4]
e) 2( ) 25 f x x = trn on [3;4]
f) 2( ) 2 5 f x x x = + trn tp xc nh.
g)4
( ) 12
f x x x
= + +
trn on [1;2]
h) 3( ) 3 sin 2 sin 1 f x x x = + trn on [0; ] i) ( ) cos 2 sin 3 f x x x = +
j) ( ) 2 sin sin 2 f x x x = + trn on [ ]32
0;
Bi 40: Tm gi tr ln nht, gi tr nh nht ca cc hm s sau y:a) 2( ) x x f x e e = + trn on [ 1;2]
b) 2( ) ( 1) x f x x e = trn on [0;2]
c) 2( ) ( 1) x f x x x e = trn on [ 1;1]
d)
2
( ) 2 2
x
f x xe x x = trn on [0;1] e) 2( ) 2( 2) 2x f x x e x x = + trn on [0;2]
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Ti liu tham kho - 22 - n tp tt nghip mn Ton
f) 2( ) ln(1 2 ) f x x x = trn on [ 2; 0]
g) 2( ) 2 4 ln f x x x x = trn on [1;2]
h) 2( ) ln( 1) f x x x = + trn on [0;2]
i) ( ) ln 2 2 f x x x x = + trn on 2[1; ]e
j) 2 2( ) 2 ln 3 f x x x x = trn on [1;2 ]e
k)2ln
( )x
f xx
= trn on [ ]31;e
l)ln
( )x
f xx
= trn on 12
[ ;e 2 ]e
Bi 41: Tm cc gi tr ca tham sm hm s sau y lun ng bina) 3 2 ( 6) 2y x mx m x = + +
b) 3 2 22( 1) (2 2) 3y x m x m m x m = + + + Bi 42: Tm cc gi tr ca tham sa hm s sau y lun nghch bin
a) 3 2( 1) (2 1) 3y x a x a x = + + + b)7
5 3
ax ay
x a
+ =
+
Bi 43: Tm cc gi tr ca m hm s sau y c cc i v cc tiua)
3 2 2
2( 1) ( 3 2) 2y x m x m m x = + + + + b)
2 2 4
2
x mx m y
x
+ =
+
c) 4 2( 1) 2 3y m x mx = Bi 44: Tm cc gi tr ca tham sm hm s:
a) 3 2 22 ( 1) ( 4) 1y x m x m x m = + + + + t cc i ti0
0x =
b) 2 3 2(2 1) (2 3) 2y m x mx m x = + + t cc tiu ti0
1x =
c)2 63
my = 3 1x mx+ + t cc tiu ti0
2x =
d) 12
y = 4 2x mx n + t cc tiu bng 2 ti0
1x =
Bi 45: Chng minh rnga) Nu (cos2 sin2 )xy e x x = + th 2 5 0y y y + =
b) Nu 4 2x xy e e= + th 13 12y y y =
c) Nu ln xyx
= th 23 0y xy x y + + =
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Dng Phc Sang - 23 - THPT Chu Vn An
PPPhhhnnn IIIIII... PPPHHHNNNGGG TTTRRRNNNH HH --- BBBT TT PPPHHHNNNGGG TTTRRRNNNH HH MMM! !! &&& LLLGGGAAARRRIIITTT
1. Phng trnh m (n gin)Cc tnh cht v lu tha cn lu : vi 0, 0a b> > v ,m n ta c
( ).
1 1
nm n m n m mn
mmnm n m n
n
n n
n n
a a a a a
aa a a
a
a aa a
+
= =
= =
= =
i i
i i
i i
( )
( ) ( )
( ) .
n
n
n n n
na a
b bn n
a b
b a
ab a b
=
=
=
i
i
i
a) Phng trnh m c bn: vi 0a > v 1a , ta c
x
a b= v nghim nu 0b logx
aa b x b= = nu 0b >
b) Phng php a v cng c s: vi 0a > v 1a , ta c( ) ( ) ( ) ( ) f x g x a a f x g x = =
c) Phng php t n s ph: Phng php gii chung:
0
Bin i phng trnh theo
( )f x
a , chng hn: 2 ( ) ( ). . 0 f x f x m a n a p+ + =
( )
( ) 1. . 0f x
f x
am a n p+ + =
1 t ( )f xt a= (km iu kin cho t) v thay vo phng trnh2 Gii phng trnh mi theo t tm nghim
0t (nu c)
3 i chiu nghim0
t tm c vi iu kin bc 1 ri tm x.
Lu 1: gp dng ( ) ( ). . 0 f x f x m a n a p+ + = , ta dng bin i
( )
( ) 1f x
f x
aa =
Lu 2: gp dng 2 ( ) ( ) 2 ( ). .( ) . 0 f x f x f x m a n ab p b+ + = , ta chia 2 v
phng trnh cho 2 ( )f xb d) Phng php lgarit ho: vi 0 1a< v 0 1b< , ta c
( ) ( ) ( ) ( )log log f x g x f x g x
a a
a b a b = =
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Ti liu tham kho - 24 - n tp tt nghip mn Ton
2. Phng trnh lgarit (n gin)Phng php chung: t iu kin xc nh ca phng trnh
Bin i phng trnh tm x(nu c) i chiu xtm c vi iu kin kt lun
Cc cng thc v quy tc tnh lgarit: vi 0 1a< v b > 0, 0 : log 1 0
a= log ( ) logn
m maa n
b b= ( 0n )
log ( )a
a = .log ( ) log loga a a
m n m n = + ( , 0m n > )
logaba b= ( )log log logma a an m n= ( , 0m n > )
log ( ) . loga a
b b = log
loglog c
c
b
a ab = ( 0 1c< )
1log logaa b b = 1loglogb
a ab = ( 1b )
a) Phng trnh lgarit c bn: vi 0a > v 1a , ta clog b
ax b x a = =
b) Phng php a v cng c s: vi 0a > v 1a , ta c log ( ) log ( ) ( ) ( )
a a f x g x f x g x = = (km iu kin ( ) 0f x > )
log ( ) ( ) ba
f x b f x a = =
Lu : Nu c ( ) 0f x > th2
log ( ) 2 log ( )n
a a f x n f x =
Nu ch c ( ) 0f x th2
log ( ) 2 log ( )n
a a f x n f x =
Bin i sau y rt dsai st(khng nn s dng):
a ra ngoi: log ( )a
f x thnh . log ( ) a f x
Tch log ( ). ( )a
f x g x
thnh log ( ) log ( )a a
f x g x +
Tch ( )( )
logf x
a g x
thnh log ( ) log ( )a a
f x g x
(ch c dng cc bin i trn khi ( ) 0, ( ) 0 f x g x > > ) Nn dng bin i di y:
a vo trong: . log ( )a
f x thnh log ( )a
f x
Nhp log ( ) log ( )a a
f x g x + thnh log ( ). ( )a
f x g x
Nhp log ( ) log ( )a a
f x g x thnh ( )( )
logf x
a g x
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Dng Phc Sang - 25 - THPT Chu Vn An
c) Phng php t n s ph:
0 Bin i phng trnh theo log ( )a
f x , chng hn:2. log ( ) . log ( ) 0a a
m f x n f x p+ + =
1 t log ( )at f x= v thay vo phng trnh.2 Gii phng trnh mi theo t tm nghim
0t (nu c)
3 T0
t t= ta gii phng trnh lgarit c bn tm x.d) Phng php m ho: vi 0 1a< v 0 1b< , ta c
log ( ) log ( )log ( ) log ( ) a b
f x g x
a a f x g x a a = =
3. Bt phng trnh m lgarit (n gin) Cng c cc cch gii nh cch gii phng trnh m, lgarit. Tuy nhin khi gii bt phng trnh m v bt phng trnh lgaritcn ch so snh c savi 1 s dng tnh ng bin, nghch binca hm s m v hm s lgarit.
Hm s m xy a= ng bin khi a> 1, nghch bin khi 0 1a< <
Hm s lgarit loga
y x= cng ng bin khi a> 1 v nghch binkhi 0 1a< <
V D MINH HOBi 1: Gii cc phng trnh sau y:
a)2 35 625x x+ = b) ( )
15 7 23
(1, 5)x
x+ = c) 12 .5 200x x+ =
Bi gii
Cu a:2 23 3 45 625 5 5x x x x + += = 2 23 4 3 4 0x x x x + = + =
hoac1 4x x = =
Vy, phng trnh cho c 2 nghim: va1 4x x= = Cu b: ( ) ( ) ( )
1 5 7 15 7 2 3 33 2 2
(1,5) 5 7 1 1x x x
x x x x+ = = = =
Vy, phng trnh cho c nghim duy nht: x= 1
Cu c: 12 .5 200 2.2 .5 200 10 100 2x x x x x x+ = = = = Vy, phng trnh cho c nghim duy nht: x= 2
Bi 2: Gii cc phng trnh sau y:a) 9 5.3 6 0
x x
+ = b)1 1
4 2 21 0x x +
+ = c) 25 2.5 5 0x x + = d) 6.9 13.6 6.4 0x x x + =
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Ti liu tham kho - 26 - n tp tt nghip mn Ton
Hng dn gii v p s
Cu a: 29 5.3 6 0 3 5.3 6 0x x x x + = + =
t 3xt = (t> 0), phng trnh trn tr thnh:
(nhan so vi )(nhan so vi )2 3 05 6 0
2 0t tt tt t
= > + = = >
3t = th 3 3 1x x= = 2t = th 33 2 log 2x x= =
Vy, phng trnh cho c 2 nghim: x= 1 v3
log 2x =
Cu b: 1 14 2 21 0x x ++ = 44
x
2.2 21 0 4 8.2 84 0x x x+ = + =
Hng dn: t 2 ( 0)
x
t t= > . p s: 2log 6x = Cu c: 2
505 2.5 5 0 5 5 0
5
x x x
x
+ = + =
Hng dn: t 5 ( 0)xt t= > . p s: 1x =
Cu d: 6.9 13.6 6.4 0x x x + = .Chia 2 v ca phng trnh cho 4x ta
c: ( ) ( ) ( ) ( )2
9 6 3 34 4 2 2
6 13 6 0 6 13 6 0x x x x
+ = + =
Hng dn: t ( )32 ( 0)x
t t= > . p s: 1x =
Bi 3: Gii cc phng trnh sau y:a)
2 2log 4 log 1 1x x + = b)
5 25 0,2log log log 3x x+ =
c) 24 82
log 2 log log 13x x x+ + = d) 233
log ( 2) log ( 4) 0x x + =
Hng dn gii v p s
Cu a:2 2
log 4 log 1 1x x + =
(1)
iu kin:4 0 4
41 0 1
x xx
x x
> > > > > . Khi ,
(1)2
log ( 4)( 1) 1 ( 4)( 1) 2x x x x = = 2( 4)( 1) 4 5 0 0x x x x x = = = hoc 5x =
So vi iu kin x> 4 ta ch nhn nghim x= 5 Vy, phng trnh cho c nghim duy nht l x= 5
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Dng Phc Sang - 27 - THPT Chu Vn An
Cu b:5 25 0,2
log log logx x+ = 13
(2) .
Vi iu kin x> 0, ( )2 11
5 5 5(2) log log log 3x x
+ =
p s: 3 3x = Cu c: 2
4 82log 2 log log 13x x x+ + = (3).
iu kin: x> 0, khi 2 12 2 23
(3) 2 log log log 13x x x + + =
p s: 8x = Cu d: 2
33log ( 2) log ( 4) 0x x + = (4).
iu kin: 22 0 2
4( 4) 0
x x
xx
> >
(I). Khi ,
23 3
(4) 2 log ( 2) log ( 4) 0x x + = 22 2
3 3 3log ( 2) log ( 4) 0 log ( 2)( 4) 0x x x x + = =
2 ( 2)( 4) 1( 2)( 4) 1
( 2)( 4) 1
x xx x
x x
= = =
p s: x= 3 v 3 2x = + Bi 4: Gii cc phng trnh sau y:
a) 22 2
log log 6 0x x = b) 22 2
4 log log 2x x+ =
c) 1 25 log 1 log
1x x +
+ = d) 2log (5 2 ) 2x x =
Hng dn gii v p s
Cu a:2
2 2log log 6 0x x = (5) iu kin: x> 0, t
2logt x= , phng trnh cho tr thnh:
2 6 0 3t t t = = hoc 2t = Vi 3t = th
2log 3 8x x= = (tho x> 0)
Vi 2t = th 22
log 2 2x x = = (tho x> 0)
Vy, tp nghim ca phng trnh (5) l: 14
{ ;8}S =
Cu b:22 24 log log 2x x+ = (6)
iu kin: x> 0, khi
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Ti liu tham kho - 28 - n tp tt nghip mn Ton
1/22 22 2 22
(6) 4 log log 2 4 log 2 log 2 0x x x x + = + =
Hng dn: t2
logt x= . p s: 12
x = v 2x =
Cu e:
1 2
5 log 1 log 1x x ++ = (7) iu kin: 0; log 1x x> v log 5x (I). t logt x= ,
(7) tr thnh 1 25 1
1 1 2(5 ) (5 )(1 )t t
t t t t +
+ = + + = +
2 5 6 0 3t t t + = = hoc 2t = Vi 3t = th log 3 1000x x= = (tho iu kin (I))
Vi 2t = th log 2 100x x= = (tho iu kin (I))
Vy, tp nghim ca phng trnh (7) l: {100;1000}S = Bi 5: Gii cc bt phng trnh sau y:
a)26 3 77 49x x+ b)( )
2 7 23 95 25
x x + +> c)4 3.2 2 0x x + <
Bi gii
Cu a: 2 26 3 7 (8) 6 3 7 27 49 7 7x x x x + + 26 3 7 2x x +
2
6 3 9 0x x + 3
2[ ;1]x (gii bng bng xt du) Vy, tp nghim ca bt phng trnh (8) l S= 3
2[ ;1]
Cu b:( ) ( ) ( )2 27 2 7 2 2(9)3 9 3 3
5 25 5 5
x x x x + + + +> > 2 7 2 2x x + + <
2 7 0 ( ; 0) (7; )x x x + < + (gii bng bng xt du)
Vy, bt phng trnh (9) c tp nghim: S= (;0)(7;+)
Cu c:4 3.2 2 0
x x
+ < (10)
t 2xt = (t> 0), (10) tr thnh: 2 3 2 0t t + < vi t> 0
Bng xt du: cho 2 3 2 0 1; 2t t t t + = = = t 0 1 2 +
2 3 2t t + + 0 0 +
Nh vy,1
2
t
t
> > <
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Dng Phc Sang - 29 - THPT Chu Vn An
Bi 6: Gii cc bt phng trnh sau y:a) 2
0,5log ( 5 6) 1x x + b) 2 2ln( 2) ln(2 5 2)x x x+ +
c) 1 13 3
2log (2 4) log ( 6)x x x+
Bi gii
Cu a: 20,5
log ( 5 6) 1x x +
iu kin: hoac2 5 6 0 2 3x x x x + > < > (I). Khi ,2 2 1
0,5log ( 5 6) 1 5 6 (0,5)x x x x + +
2 5 4 0 1 4x x x +
Kt hp vi iu kin (I) ta nhn cc gi tr: [1;2) (3;4]x Vy, tp nghim ca bt phng trnh l: [1;2) (3;4]S =
Cu b: 2 2ln( 2) ln(2 5 2)x x x+ +
iu kin:hien nhien
2
2
2 5 2 0
2 0 :
x x
x
+ > + >
hoac12
x 2x < > (I)
Khi , 2 2 2 2ln( 2) ln(2 5 2) 2 2 5 2x x x x x x + + + + 2 5 0 0 5x x x
Kt hp vi iu kin (I) ta nhn cc gi tr: 12
[0; ) (2;5]x
Vy, tp nghim ca bt phng trnh l: 12
[0; ) (2;5]S =
Cu c: 1 13 3
2log (2 4) log ( 6)x x x+
iu kin:
hoac2 2 36 0322 4 0
x xx x
xxx
< > > > > + >
Vi iu kin x> 3 ta c
1 13 3
2 2log (2 4) log ( 6) 2 4 6x x x x x x + +
2 3 10 0 2 5x x x Kt hp vi iu kin x> 3 ta nhn cc gi tr 3 5x< Vy, tp nghim ca bt phng trnh l:
(3;5]S
=
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Ti liu tham kho - 30 - n tp tt nghip mn Ton
BI TP V PHNG TRNH MBi 7: Gii cc phng trnh sau y:
a) 27 8.7 7 0x x + = b) 22.2 2 1 0x x+ =
c) 9 3 6 0x x = d) 25 2.5 15 0x x+ =
e) 2 12 2 6x x+ = f) 2 38 2 56 0x x =
g) 33 3 12x x+ = h) 32 2 2 0x x + =
i) 2 3 25 5 20x x = j) 17 2.7 9 0x x+ =
k) 2 24. 3x xe e = l) 16 2.6 13 0x x+ + =
m)3.4 2.6 9x x x = n) 2 125 10 2x x x++ =
o) 25 15 2.9x x x
+ = p) 5.4 2.25 7.10 0x x x
+ = q) 6 33. 2 0x xe e + = r) 4 12 15.4 8 0x x+ =
s) 2 15 5.5 250x x + = t) 2 13 9.3 6 0x x+ + =
u) 2 6 72 2 17x x+ ++ = v) 1 1 12 (2 3 ) 9x x x x + = Bi 8: Gii cc phng trnh sau y:
a) 2 5 2 32 2 12x x+ ++ = b) 4 2 12 2 5 3.5x x x x + + ++ = +
c)2 1 2
3 3 108x x
+ = d)2 2
5 7 .17 7 5 .17x x x x
+ = + e) 1 22 .5 0,2.10x x x = f)
25 5 11 1 2 212 .4 48.3x x x x + =
g) 3 1 3 28.4 2x x = h) 3 3 1 12 .3 2 .3 192x x x x + =
i)2 2 13 .2 72x x x x + = j) 1 2 3(0,25) 2 0,125.16x x =
Bi 9: Gii cc phng trnh sau y:a) 13.2 4 1 0x x++ = b) 2 4 15 110.5 75 0x x+ + =
c) ( ) ( ) 15 7 231, 5xx + = d) ( )
52 22 16
9(0, 75) 0xx x =
e) 2 1 23 3 108x x + = f) 2( 1)16 2 12 0x x++ =
g) 4.9 12 3.16 0x x x+ = h) 4 8 2 53 4.3 27 0x x+ + + =
i) 13 (3 30) 27 0x x+ + = j) 3 2 1 32 2 2 0x x x+ + =
k) 2 22 9.2 2 0x x+ + = l) 1 3 21 3.2 2 0x x + =
m)2 1 2
3 2.3 5 0x x
+ = n) 4.9 12 3.16 0x x x
+ = o)
2 222 2 3x x x x + = p) 4 2 22.16 2 4 15x x x =
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Dng Phc Sang - 31 - THPT Chu Vn An
q) ( ) ( )2 33 24. 2. 6 0x x
+ = r) ( ) ( )2 3 2 3 4x x
+ + =
s) 12 .4 64 5 0x x x + = t) 14 4 .4 3 0x x x+ + =
u)
1
36 3 .2 4 0
x x x+
= v)
1
4 2 .4 3 0
x x x
= BI TP V PHNG TRNH LGARITBi 10: Gii cc phng trnh sau y:
a) 2log( 6 5) log(1 )x x x + = b) 4 22
ln . log ( 2 ) 3 lnx x x x =
c) 17
27
log ( 2) log (8 ) 0x x+ + = d) 13
23
log ( 10) log (3 ) 0x x + =
e) ln(4 4) ln( 1) lnx x x = f)22
log ( 1) log (7 )x x =
g) 2 4log 2 log ( 1) 1x x + + = h) 13
3log ( 2) log ( 4) 1x x =
i)2 2
log ( 1) log (2 11) 1x x = j)2 4 0,5
log (2 ) log logx x x+ =
k)2 0,5
log ( 3) log ( 1) 3x x + = l)5 0,25
log log log 2x x x+ =
m)3 9 27
log log log 11x x x+ + = n) 4 3log log(4 ) 2 logx x x+ = + Bi 11: Gii cc phng trnh sau y
a)25 5log 4 log 3 0x x + = b)
22 22 log log 1 0x x+ =
c) 25 0,2
log log 12 0x x+ = d) 2ln ln( ) 1 0x ex =
e) 22 0,5
log 5 log 4 0x x+ + = f) 22 0,5 2
3 log log log (2 )x x x =
g) ( )22 4 8log 6 log 7xx = h) 2
0,2 5log 5 log 6 0x x+ + =
i) 2 2log 3 log log 4x x x = j) 2log (10 ) 9 log(0,1. )x x=
k) 3log log 9 3xx + = l) 3log 27 3 log 8x x = m)
22 log 2 log 5
xx+ = n) ( )6 62 log 5 log 6
xx
x =
Bi 12: Gii cc phng trnh sau ya) 2
3 3log ( 5) log (2 5)x x x = + b) log (2 ) log (10 3 )x x
=
c) 3 3log log4 5.2 4 0x x + = d) 2log (10 ) 3 log 1 0x x =
e) 55log ( 2) log (4 5)x x+ = + f)2
3 3log (3 ) log 1 0x x+ = g) 2
2 0,52log 3 log log 2x x x+ + = h) 2 3log log 2 0x x + =
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Ti liu tham kho - 32 - n tp tt nghip mn Ton
i) log 1 log 2 1log 2 log 1 2
x x
x x
+ +
= j) 2 84 16
log log (4 )
log (2 ) log (2 )
x x
x x=
k) 13 3log (3 1). log (3 3) 6x x+ = l)
5 5log ( 2) log ( 6)x x x+ = +
m) 3log(10 ).log(0,1. ) log 3x x x= n)
4 22log 4 log log (4 ) 12x x x+ + =
o) 2 14 22log ( 2) log (3 1) 1x x + =
p)2 2
1log log ( 1)( 4) 2
4
xx x
x
+ + = +
BI TP V BT PHNG TRNH M - LGARITBi 13: Gii cc bt phng trnh sau y
a)22 3(0, 5) 2x x b)2 2 3 0x x+ < c)
2 32 4x x + <
d) 2 13 3 28x x+ + e)4 3.2 2 0x x + > f)2
3 9x x < Bi 14: Gii cc bt phng trnh sau y
a) 2 6 72 2 17x x+ ++ > b) 2 3 25 2.5 3x x c)4 2 3x x> +
d) 4 4 2 22.2 2 4 15x x x e)5.4 2.25 7.10x x x+ f) 14 16 3x x+ Bi 15: Gii cc bt phng trnh sau y
a)2 2
log ( 5) log (3 2 ) 4x x+ b) 13
52
log log 3 x
x >
c) 28 8 32 log ( 2) log ( 3)x x > d) 1
3
3 1log 1
2
x
x
>
+
e)4 4
log ( 7) log (1 )x x+ > f) 22 2
log log 0x+ Bi 16: Gii cc bt phng trnh sau y
a) 1 12 2
2log (5 10) log ( 6 8)x x x+ < + +
b)2 2
log ( 3) log ( 2) 1x x + c) 1 12 2
log (2 3) log (3 1)x x+ > +
d)0,2 0,2
log (3 5) log ( 1)x x > + e)3 3
log ( 3) log ( 5) 1x x + <
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Dng Phc Sang - 33 - THPT Chu Vn An
PPPhhhnnn IIIIIIIII... NNNGGGUUUYYYNNN HHHM MM --- TTTCCCH HH PPPHHHNNN VVV (((NNNGGG DDD***NNNGGGI. TM TT CNG THC V PHNG PHP GII1. Bng cng thc nguyn hm v nguyn hm m rng
11
2 2
1. .( )1
. ( ) .1 1
1 1 ln. ln .
1 1 2. 2 .
1 1 1 1 1. .
( )
. .ax b
x x ax b
dx x C a dx ax C
ax bxx dx C ax b dx C
aax b
dx x C dx C x ax b a
ax bdx x C dx C
ax ax b
dx C dx C x a ax bx ax b
ee dx e C e dx
a
++
++
= + = ++
= + + = ++ +
+= + = +
++
= + = ++
= + = +++
= + =
i i
i i
i i
i i
i i
i i
2 2
2 2
sin( )cos . sin cos( ).
cos( )
sin . cos sin( ).tan( )1 1
. tan .cos cos ( )
cot( )1 1. cot .
sin sin ( )
C
ax bx dx x C ax b dx C
aax b
x dx x C ax b dx C a
ax bdx x C dx C
ax ax b
ax bdx x C dx C
ax ax b
+
+= + + = +
+
= + + = ++
= + = ++
+= + = +
+
i i
i i
i i
i i
2. Cng thc tch phnVi ( )F x l mt nguyn hm ca hm s ( )f x trn on [ ; ]a b th
( ) ( ) ( ) ( )b b
aaf x dx F x F b F a = =
3. Phng php i bin s (loi 2): xt ( ) . ( ).b
aI f t x t x dx =
1 t ( )t t x= ( ).dt t x dx = (v 1 s biu thc khc nu cn)2 i cn: ( )x b t t b= =
( )x a t t a = =
3 Thay vo:( )
( ).( )t b
t aI f t dt = v tnh tch phn mi ny (bin t)
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Ti liu tham kho - 34 - n tp tt nghip mn Ton
Vi dng tch phn i bin thng dng:Dng tch phn Cch t c im nhn dng
( )
( )
t xdx
t x
( )t t x= mu
( ). ( )t xe t x dx ( )t t x= m( )( ) . ( ).f t x t x dx ( )t t x= ngoc
( )n ( ) . ( )f t x t x dx n ( )t t x= cn
( )ln dxf xx
lnt x= ln x
(sin ).cos f x xdx sint x= cos .x dx i km biu thc theo sin x (cos ).sin f x xdx cost x= sin .x dx i km biu thc theo cos x
2(tan )
cos
dxf x
x tant x= 2cos
dx
xi km biu thc theo tan x
2(cot )
sin
dxf x
x
cott x= 2sin
dx
x
i km biu thc theo cotx
( ).ax ax f e e dx axt e= axe dx i km biu thc theo axe i khi thay cch t ( )t t x= bi . ( )t m t x n = + ta s gp thun li hn
4. Phng php tch phn tng phn
( ). . .b bb
aa au dv u v v du =
Vi dng tch phn i bin thng dng:Vi ( )P x l mt a thc, ta cn ch cc dng tch phn sau y
( ).sin .P x ax dx , ta t( )
sin .
u P x
dv ax dx
= =
( ).cos .P x ax dx , ta t( )
cos .
u P x
dv ax dx
= =
( ). .axP x e dx , ta t ( ).axu P xdv e dx = =
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Dng Phc Sang - 35 - THPT Chu Vn An
. sin .axe bx dx , ta t sin .axu e
dv bx dx
= =
(khong co )
( ). ln . ,n f x x dx
dx
x
ta t ln( ).
n
u xdv f x dx
= =
5. Tnh din tch hnh phngCho hai hm s ( )y f x= v
( )y g x= u lin tc trn on
[ ; ]a b ,Hl hnh phng gii hn
bi cc ng: 1 2( ) : ( ),( ) : ( ),C y f x C y g x x a = = = v x b= Khi , din tch ca hnh phngHl: ( ) ( )
b
aS f x g x dx =
Lu 1: nu2
( )C l trc honh th ( ) 0g x = v ( )b
aS f x dx =
Lu 2: Khi tnh tch phn ( )b
as x dx ta cn lu nh sau:
Nu ( ) 0, [ ; ]s x x a b th
( ) ( ).b b
a as x dx s x dx =
Nu ( ) 0, [ ; ]s x x a b th
( ) ( ).b b
a as x dx s x dx =
Nu ( )s x khng c nghim trn khong ( ; )a b th
( ) ( ).b b
a a
s x dx s x dx =
Nu ( )s x c nghim 1 2 nc c c< <
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Ti liu tham kho - 36 - n tp tt nghip mn Ton
V D MINH HO
Bi 1: Tnh 30 2
3
1
xA dx
x=
+ 2
3
1 cos
sin (1 cos )
xC dx
x x
=
+
22
13 . .xB x e dx
=
4
2ln 1
. lnxD dx
x x+=
Bi gii
Cu a:3
0 2
3
1
xA dx
x=
+ t
2 2 21 1t x t x = + = + 2 . 2 . . .t dt x dx t dt x dx = =
i cn: 3x = 2t = 0x = 1t =
Vy, ( )2 2 211 1
3. 3. 3 6 3 3tdtA dt t t
= = = = =
Cu b:22
13 . .xB x e dx
= t 2t x= 122dt xdx xdx dt = =
i cn: 2x = 4t = 1x = 1t =
Vy, ( )44 43 3 3
2 2 21 1
3 .
2
tte dtB e e e = = =
Cu c: 2 23 3
2
1 cos sin
sin (1 cos ) (1 cos )
x xC dx dx
x x x
= =
+ +
t 1 cos sin .t x dt x dx = + = sin .x dx dt =
i cn:2
x = 1t =
3x = 3
2t =
Vy, ( )33
2232
1
12 21 11 . t
dtC dt
t t= = = ( )2 1 13 1 3= =
Cu d:4
2
ln 1
. ln
xD dx
x x
+= t
1lnt x dt dx
x= =
i cn: 4x = 2ln2t = 2x = ln 2t =
Vy, ( )ln 4ln 4 ln 4
ln2 ln2 ln2
1 11 ln
tD dx dt t t
t t
+ = = + = +
( ) ( )ln 4 ln ln 4 ln 2 ln ln 2 ln 4 = + + =
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Dng Phc Sang - 37 - THPT Chu Vn An
Bi 2: Tnh cc tch phn sau y: 20
( 1) sinE x xdx =
2
13 . xF x e dx
=
22
1(3 1) ln .G x x dx =
Bi gii
Cu e: 20
( 1) sinE x xdx =
t1
sin cos
u x du dx
dv xdx v x
= = = =
Suy ra, ( ) ( )22 20 00
( 1)cos cos 0 1 sinE x x xdx x = + = +
21 sin sin 0 0= + =
Cu f:2
13 . xF x e dx
= t 3 3x xu x du dx dv e dx v e
= = = =
Nh vy, ( ) ( )2 22
2 1
11 13 . 3 6 3 3x x xF x e e dx e e e
= = +
2 2 1 2 2 23 3 3 66 3( ) 6 3 3e e e e e e e e e e
= + = + + = +
Cu g:2 2
1(3 1) ln .G x x dx = t 2
3
1ln(3 1)
u x du dx x
dv x dx v x x
= = = =
( ) ( )2 223 2 31 4
3 311 1ln ( 1). 6 ln 2 6 ln 2G x x x x dx x x = = =
Bi 3: Tnh cc tch phn sau y2
1
1xH x e dx x
=
22
0 ( 1).I x x xdx = + + 3
21
2 1e t tJ dt
t
+= 20 (1 2 sin )sinK a ada
= +
Bi gii
Cu h:2 2 2 2
1 1 1 1
1( 1) 1.x x xH x e dx xe dx xe dx dx
x
= = =
Xt 21 1:xH xe dx = t x xu x du dx dv e dx v e = = = =
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Ti liu tham kho - 38 - n tp tt nghip mn Ton
( ) ( )2 22
2 21 11 1
. 2x x xH xe e dx e e e e = = = =
Xt ( )2 2
2 111 2 1 1H dx x = = = =
Vy, 21 2 1H H H e = =
Cu i:2 2 2
2 2 2
0 0 0( 1). . 1.I x x x dx x dx x xdx = + + = + +
Xt ( )22 2 31 8
1 3 30 0I x dx x = = =
Xt2
22 0
I 1.x xdx = + . t 2 1t x tdt xdx = + =
i cn: 2x = 5t = 0x = 1t =
( )55 5 2 31
2 31 1 1.I t tdt t dt t = = = 5 5 13
=
Vy, 5 5 71 2 3I I I+= + =
Cu j:3 2
2 22 1 2 1 1
1 1 2 12 ln
e
e et t tt tt t
J dt t dt t + = = + =
( ) ( )2 21 1 1 1 32 2 1 2 22 ln 2 ln 1e ee ee= = Cu k: 2 2 2
0 0(1 2 sin )sin (sin 2 sin )K a ada a a da
= + = + 2
0(sin 1 cos 2 )a a da
= + ( ) 2sin22 0cosaa a
= +
( ) ( )sin sin 02 2 2 2 2cos cos 0 0 1= + + = +
Bi 4: Tnh din tch hnh phng gii hn bi cc ng sau y:a) 3 3 2y x x= + , trc honh, 1x = v 3x =
b) 24y x= v 2 42y x x=
c) 3 2y x x= v tip tuyn ca n ti im c honh bng 1
d) 3y x x= v 2y x x=
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Dng Phc Sang - 39 - THPT Chu Vn An
Hng dn gii v p s
Cu a: Xt3
3( ) 3 2 ( ) ( ) 3 2( ) 0
f x x x f x g x x x
g x
= + = + =
Din tch cn tm l 2 31
3 2S x x dx
= + Bng xt du ca 3 3 2x x + trn on [ 1;2]
x 1 1 23 3 2x x + + 0 +
Vy, ( )2
3
13 2S x x dx
= + ( )
4 22
3 214 2 4
12x x x
= + =
Cu b: Xt2
4 22 4
( ) 4( ) ( ) 3 4
( ) 2
f x x f x g x x x
g x x x
= = =
Cho 4 23 4 0x x = 2x =
Din tch cn tm l2
4 2
23 4S x x dx
=
Bng xt du ca 4 23 4x x trn on [ 2;2]
x 2 24 23 4x x
( )22 4 2 5 31 96
5 52 2( 3 4) 4S x x dx x x x
= = =
Cu c: HD: vit phng trnh tip tuyn tho (p s: 2y x= + )
Xt3
3( ) 2 ( ) ( ) 3 2
( ) 2
f x x x f x g x x x
g x x
= =
= +
Cho 3 3 2 0 1x x x = = hoc 2x =
Din tch cn tm l:2
3
13 2S x x dx
=
Bng xt du ca 3 3 2x x trn on [ 1;2] x 1 2
3 3 2x x
( )22 3 4 2 271 34 2 41 1( 3 2) 2S x x dx x x x = = =
1
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Ti liu tham kho - 40 - n tp tt nghip mn Ton
Cu d:Xt3
3 22
( )( ) ( ) 2
( )
f x x x f x g x x x x
g x x x
= = + =
Cho 3 2 2 0 2; 0; 1x x x x x x + = = = = .
Din tch cn tm l1 3 2
22S x x x dx
= +
HD: xt du 3 2 2x x x+ v a n cng thc0 1
3 2 3 2
2 0( 2 ) ( 2 )S x x x dx x x x dx
= + +
( ) ( )0 1
4 3 2 4 3 2 371 1 1 14 3 4 3 122 0
x x x x x x
= + + =
Bi 5: Tnh th tch vt th trn xoay sinh ra khi quay hnh (H)quanhtrc Oxbit (H) gii hn bi: siny x= ,Ox, 0x = v 3
2x =
Bi gii
Ta c, ( ) sin f x x = . Xt on [ ]32
0;
Th tch cn tm l:32 2
0(sin )V x dx
= 3 3 32 2 22
0 0 0
1 cos 2 1 cos 2sin2 2 2
x xV xdx dx dx
= = =
( ) ( )3
221 1 3 1 32 4 4 4 40
sin 2 sin 3 .0x x
= = =
BI TP V TCH PHNBi 6: Tnh cc tch phn sau ya)
12
0
.(2 1)x x dx
b)
ln2
0
(3. 5)x xe e dx
c)
13
1
(2 3 )x dx
d)2
1
1 tte tdt
t
+ e)
2
1
(1 ) x
x
x e xdx
xe
+ f)
23
1
3 2t tdt
t
+
g) ( )22
11 t
t dt h) ( )21
22
x x dx x
+ i)1
3
0(1 )x x dx
j) 46
cos4 .cos3x xdx
k) 64
sin 3 . sin .t t dt
l)4 2
0tan xdx
m) 120
1 .cos
xx ee dxx
+ n)2 1ln2
0
1xx
e dxe
+ + o)2
01 x dx
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Dng Phc Sang - 41 - THPT Chu Vn An
p)32
1
2 5t tdt
t
q)
22
0
3 1
1
x xdx
x
+ r) 12
1 3 1
( 1)
xdx
x x
++
m) 3
6
2 2
2
tan cos
sin
x xdx
x
n) 3
20
2 cos2 1
cos
xdx
x
o) 4 2
0
sin .x dx
Bi 7: Tnh cc tch phn sau ya) 2
0
sin
1 3 cos
xdx
x
+ b)2
21
1
2 3
xdx
x x
c)
21 1
0. xx e dx
d)1/2
21
xedx
x e) 2
62
cos
(1 sin )
xdx
x
+f)
20
41 (1 )
xdx
x
g) 20sin .
8 cos 1
x dx
x
+ h)19
0 3 2
3
8
xdx
x + i)2
1
1e ln xdxx
+
j)1
1
(1 ln )e
e
dxx x k)
3
1 . 4 ln
e dx
x x
l)1
ln .
.(ln 3)e
e x dx
x x +
m)1
2012
0( 1)x x dx n)
12
01x x dx + o)
73
0. 1x x dx +
p)2
2
3
sin .cos .x x dx
q) 40
sin2
.cos2
x
e xdx r)0
5 4 .x x dx s)
22
sin2
1 cos
xdx
x
+t)
1
2 20
4
(2 1)
xdx
x +u)
ln 3
0 1 xdx
e+
Bi 8: Tnh cc tch phn sau ya)
1
0( 1) xx e dx + b)
1
0(2 1) xx e dx c)
12 1
0. xx e dx
d)
ln 5
ln2 2 ( 1)x
x e dx e)ln2
0 ( 1)x
x e dx
f)2
x x dx 0 2 .cos .
g) 4
0(2x 1)cos
xdx h)0
(1 )cosx xdx
i) 20 2 .sinx xdx
j) 40
(x 1)sin2
xdx+ k) 40 x xdx sin2
l) 1 ln .e
x dx
m)1
2 .(ln 1)e
x x dx n)3
22 ln( 1)x x dx o)
2
21
ln xdx
x
p)3
2 2
0( 1). xx e dx + q) 40 sin
xe xdx
r)4
1
xe dx
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Ti liu tham kho - 42 - n tp tt nghip mn Ton
Bi 9: Tnh cc tch phn sau ya)
1
0(3. 5 )x xe e x dx b) ( )0 x x cos
x dx+ c)2
2
0( )xx x e dx +
d)
2
1
lnx x
dxx
+
e)4
1
xx e
dxx
+
f) 211e lnx x
dxx
+
g) ( )1
ln 1e
x x dx + h) 40 (x x xdx cos )sin
+ i)2
1( 2 )xx xe dx +
j)1
0
1
1
x
x
xe xdx
e
+ +
+k) 2
0
1 sin
1 cos
xdx
x
+ l)
2
21
( 1). lnx xdx
x
Bi 10: Tnh cc tch phn sau y1) ( )0 2 11 xx ee dx 2)
2
1 ( 1)dx
x x + 3)6
0cos
2 sin 1xdxx
+
4)1
03 1.x dx+ 5)
2
1(2 1) ln .x x dx + 6) 1 ln( 1)
e
x dx+
7)22
1
1 ln xdx
x
+ 8)
4
1
lne .x dx
x 9)2 22
1
lnx xdx
x
+
10)
1
0
2 1
1
x
dxx
+ 11)4
1 ( 2)
dx
x x + 12)32 2
0 2 1
x dx
x +
13) 4tan
20 cos
xe dx
x
14) 20
cos sin
1 cos
x xdx
x
+ 15)
2ln2
30 ( 4)
x
x
e dx
e +
16)0
ln63.x xe e dx + 17) 0 ( cos )
xx e
x dx+ x xdx 18) 0 2 sin
19) 34
30
cos sin
cos
x xdx
x
+
20)
21 (ln 1)
e dx
x x +21)
2
1
ln .
(ln 2)
e x dx
x x +
22) 2 20
sin 2 .sin .x x dx
23) 2 20
sin .cosx xdx
24)1
0(4 1) xx e dx +
25)2
1
ln 1e x xdx
x
+ 26) 20
sin2 .
1 cos
x dx
x
+ 27)2
0
sin 2 .
3 sin 1
x dx
x
+
28)0
(1 cos ) cos .x x dx
29) ( )2
04 1x x dx + 30)
1
0( 3)xxe dx +
31) ( cos 2)x x dx
32) 1 ( ln 2)
ex x x dx + 33)
21 3
0
xx e dx
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Dng Phc Sang - 43 - THPT Chu Vn An
BI TP V NG DNG HNH HC CA TCH PHNBi 11: Tnh din tch hnh phng gii hn bi cc ng sau y
a) 3 21 23 3
y x x= + , trc honh, x= 0 v x= 2.
b)2
1, 1, 2y x x x = + = = v trc honh.c) 3 12y x x= v 2y x= .
d) 2 2y x x= + v 2y x+ = .
e) 3 1y x= v tip tuyn ca n ti im c tung bng 2.
f) 3 3 2y x x= + v trc honh.
g)2
2y x x= v2
4y x x= + h) 2 2y x x= v y x=
i) 3 2y x x= v ( )91y 1x=
j) ( ) : 1 , 1C xy x x = + = v tip tuyn vi ( )C ti im ( )322; .
k)3 1
, , 01
xy Ox x
x
+= =
l) 1ln , ,e
y x x x e = = = v trc honh.
m)ln
1x
y xx
= + , 1y x= v x e=
Bi 12: Tnh th tch cc vt th trn xoay khi quay cc hnh phng giihn bi cc ng sau y quanh trc km theoa) 2 4 ,y x x= trc honh, 0, 3x x= = ( l trc honh)
b) cos ,y x= trc honh, 0,x x = = ( l trc honh)c) tan ,y x= trc honh,
40,x x = = ( l trc honh)
d) ,xy e x= trc honh v 1x = ( l trc honh)
e)2
,2
yx
=
trc honh, 0, 1x x= = ( l trc honh)
f) 22 , 1y x y= = ( l trc honh)
g) 22y x x= v y x= ( l trc honh)h) 3 2 1y x= + , 3y = v trc tung ( l trc tung)
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Ti liu tham kho - 44 - n tp tt nghip mn Ton
BI TP V NGUYN HM
Bi 13: Chng minh rng hm s 2( ) ( 1)xF x e x = + l mt nguyn hmca hm s 2( ) ( 1)x f x e x = + trn .
Bi 14: Chng minh rng hm s ( ) ln 3F x x x x = + l mt nguynhm ca hm s ( ) ln f x x = trn .
Bi 15: Chng minh rng 4 4( ) sin cosF x x x = + v 1 cos 4( )4
xG x
= l
nguyn hm ca cng mt hm s vi mi xthuc
Bi 16: Tm gi tr ca tham sm 3 2( ) (3 2) 4 3F x mx m x x = + + + l mt nguyn hm ca hm s 2( ) 3 10 4 f x x x = + trn
Bi 17: Tm a,b v c 2( ) ( ) xF x ax bx c e = + + l mt nguyn hm cahm s ( ) ( 3) x f x x e = trn
Bi 18: Tm nguyn hm ( )F x ca hm s ( ) cos (2 3 tan ) f x x x = bitrng ( ) 1F =
Bi 19: Tm nguyn hm ( )F x ca hm s 21 2( ) xf xx
+= tha mn iu
kin ( 1) 3F = .
Bi 20:Tm nguyn hm ( )F x ca hm s2
1 ln( )
xf x
x
+= tha mn
iu kin ( ) 0F e = .
Bi 21: Tm nguyn hm ( )F x ca hm s 2( ) (2 ) f x x x = tha mniu kin ( 1) 3F = .
Bi 22: Tm nguyn hm ( )F x ca hm s 2(1 2 )( ) xf xx
= tha mn
iu kin ( 1) 1F = .
Bi 23: Tm nguyn hm ( )F x ca hm s ( ) (4 1) x f x x e = + tha mniu kin (1)F e= .
Bi 24: Tm nguyn hm ( )F x ca hm s (1 ln )( ) xx x ex
f x+
= tha mn
iu kin (1)F e= .
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Dng Phc Sang - 45 - THPT Chu Vn An
PPPhhhnnn IIIVVV... SSS PPPHHH(((CCC
1. Cc khi nim v php ton lin quan n s phc n v o i: 2 1i = i 3i i= i 4 1i =i S phc z a bi = + l s c phn thc l a v phn o b Mun ca s phc z a bi = + l: 2 2z a b= + S phc lin hp ca s phc z a bi = + l: z a bi =
Hai s phc bng nhau:a c
a bi c di b d
=+ = + =
Php cng hai s phc: ( ) ( ) ( ) ( )a bi c di a c b d i + + + = + + +
Php tr hai s phc: ( ) ( ) ( ) ( )a bi c di a c b d i + + = + Php nhn hai s phc: ( ).( ) ( ) ( )a bi c di ac bd ad bc i + + = + +
Php chia hai s phc: 1 1 2
2 2 2
.
.
z z z
z z z= (nhn c t ln mu cho
2z )
S phc nghch o ca zl: 1.z
z z z=
Mi s thc am c 2 cn bc hai phc l: .a i
Ch :s phc ch c phn o (phn thc bng 0) gi l s thun o2. Gii phng trnh bc hai h s thc ( < 0) trn tp s phc
Cho phng trnh bc hai va2 0 ( , , 0)az bz c a b c a + + =
Tnh 2 4b ac = v ghi kt qu di dng 2( . )i Kt lun phng trnh c 2 nghim phc:
1z =
2
b i
a
v2
z =2
b i
a
+
Lu :
Ch c dng cng thc nghim nu trn khi < 0 Trng hp 0 ta gii pt bc hai trn tp s thc (nh trc).
Khi gii phng trnh trng phng trn C, ta t 2t z= (khng cn
iu kin cho t)Nu dng bit thc th cng thc tm hai nghim phc l
1z = b i
a v
2z = b i
a +
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Ti liu tham kho - 46 - n tp tt nghip mn Ton
V D MINH HOBi 1: Thc hin cc php tnh
a)(2 4 )(3 5 ) 7(4 3 )i i i+ + b) 2(3 4 )i c) 23 2
i
i
++
Bi giiCu a: 2(2 4 )(3 5 ) 7(4 3 ) 6 10 12 20 28 21i i i i i i i + + = + +
6 10 12 20 28 21 54 19i i i i = + + + = Cu b: 2 2(3 4 ) 9 24 16 9 24 16 7 24i i i i i = + = =
Cu c:2
2 2 2
(2 )(3 2 )2 6 4 3 2 6 2 8 13 2 (3 2 )(3 2 ) 13 133 4 3 4
i ii i i i i
i i i ii
+ + + ++ + +
= = = =
Bi 2: Tm mun ca s phc sau ya) 23 2 (1 )z i i= + + + b) 3
(1 )(2 )i
i iz +
+ =
Bi gii
Cu a: 23 2 (1 ) 3 2 3 221 2 1 2 1z i i i i i i i = + + + = + + + + = + + + 2 2 2 23 4 3 4 5z i z a b = + = + = + =
Cu b:2
3 3 3 3
(1 )(2 ) 2 2 1 32 2
1 1i i i i
i i i i i i i i
z z+ + + +
+ + + + +
= = = = = =
Bi 3: Tm s phc nghch o ca s phc: 2(1 ) (2 )z i i= + Bi gii
2 2 2(1 ) (2 ) (1 2 )(2 ) ( 2 )(2 ) 4 2 2 4z i i i i i i i i i i = + = + + = + = =
Suy ra2
2 4 2 4 2 41 1 1 12 4 (2 4 )(2 4 ) 20 10 54 16
i i i
z i i i ii+ + +
+ = = = = = +
Bi 4: Gii phng trnh sau trn tp s phc: 2 3 5 4iz z i + = + Bi gii
2 3 5 4 5 2 3 4 (5 2 ) 3 4iz z i z iz i i z i + = + = = 2
2 2
(3 4 )(5 2 ) 15 6 20 83 4 23 145 2 (5 2 )(5 2 ) 29 295 4
i i i i ii
i i i iz i
+ + +
= = = =
Bi 5: Gii cc phng trnh sau y trn tp s phc:a) 2 2 0z z + = b) 4 22 3 0z z+ = c) 3 1 0z + =
Bi giiCu a: 2 22 0 2 0z z z z + = + = (1)
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Dng Phc Sang - 47 - THPT Chu Vn An
Ta c, 2 21 4.1.2 7 0 ( 7. )i = = < = Vy, phng trnh (1) c 2 nghim phc phn bit
1 7 711 2 2 2
iz i= = v 1 7 712 2 2 2iz i+= = +
Cu b: 4 22 3 0z z+ = (2)
t 2t z= , phng trnh (2) tr thnh:
2 321
03
tt t
t
=+ = = . T ,
2
2
11
3.3
zz
z iz
= = = =
Vy, phng trnh (2) c 4 nghim phc phn bit :
1 2 3
1, 1, 3.z z z i = = = v4
3.z i=
Cu c: 3 (3) 2 2 (*)1
1 0 ( 1)( 1) 01 0
zz z z z
z z
= + = + + = + =
Gii (*) : ta c 2 2( 1) 4.1.1 3 0 ( 3 )i = = < =
(*) c 2 nghim phc phn bit: 1 31 2iz += ; 1 32 2
iz =
11
Vy, phng trnh (3) c 3 nghim phc phn bit
z = , 312 2 2z i= + v31
3 2 2z i=
Bi 6: Tm mun ca s phc zbit:a) 3 (3 )(1 ) 2iz i i + + = b) 5 11 17iz z i + =
Bi gii
Cu a: 23 (3 )(1 ) 2 3 3 3 2iz i i iz i i i + + = + + =
3 3 3 1 2 3 2 2iz i i iz i + + + = = 2 2
3
i
i
z =
2 23 3
z i = + 2 2z a b = + = ( ) ( )2 2
2 22 23 3 3
+ =
Cu b: Vi ( , )z a bi a b= + ta c z a bi = , do 5 11 17 ( ) 5( ) 11 17iz z i i a bi a bi i + = + + =
2 5 5 11 17 (5 ) ( 5 ) 11 17ia bi a bi i a b a b i i + + = + =
2 25 11 3
3 4 3 4 55 17 4
a b a
z i za b b
= = = + = + = = =
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Ti liu tham kho - 48 - n tp tt nghip mn Ton
III. BI TP V S PHCBi 7: Thc hin cc php tnh
a) 2(1 )i+ b) 2(3 4 )i c) 2( 2 )i + d) 3(2 3 )i+
e) 3(1 3 )i f) 2012(1 )i g) 2012(1 )i+ h) 2012(1 3 )i
i) 2 33
i
i
++
j) 4 21
i
i
+
k) 2 4ii
+ l) 12 i
m)2
1
1
i
i
+n)
51
1
i
i
+ o)
2
2
(2 )ip)
(2 1)
1
i i
i
+
Bi 8: Xc nh phn thc, phn o v mun ca cc s phc sau y:a)(2 4 )(3 5 ) 7(4 3 )i i i+ + b)(1 4 )(2 3 ) 5( 1 3 )i i i +
c) 2(1 2 ) (2 3 )(3 2 )i i i + d) 2(2 3 ) (1 3 )(5 2 )i i i + e) 2 2(1 2 ) (1 2 )i i+ + f) 2 2(1 3 ) (1 3 )i i+
g)5
(4 5 ) (4 3 )i i + + h)3
(5 ) (2 7 )i i +
i)(2 ) (1 )(4 3 )
3 2
i i i
i
+ + + +
j)(2 ) (1 )(1 3 )
3 9
i i i
i
+
k)(3 4 )(1 2 )
4 31 2
i i
ii
++ l)
(2 3 )(1 2 )(2 4 )1
i i
ii
+ + +
Bi 9: Gii cc phng trnh sau trn tp s phc:a) 3 8 5 4z i i+ = + b) 22 (2 ) 2 3iz i i + = +
c) (3 ) (1 )(4 2 )i z i i = + d) 2(1 ) (1 ) 2 3i z i i + + =
e)2 1 3
1 2
i iz
i i
+ +=
+f)
2 1 3
1 2 2
i iz
i i
+ =
+ +
g) (2 ) 3 2i z i i + = + h)2 . 1 5. 2i z z i = i) 2 3 5 4iz z i + = + j) 3 . 5 3z i z i = k) 2 6 2z z i+ = + l) 3 7 5iz z i + = + m)3 2 5 2z z i+ = + n) . 2 2 5i z z i + =
Bi 10: Tnh z , bit rng a) 2(1 2. )z i= + b) 34
(1 )
(1 )
i
iz
+
=
Bi 11: Tm s phc nghch o ca cc s phc sau y:a) 3 4z i= b) (4 )(2 3 )z i i= + c) 2(2 )z i i=
Bi 12: Cho1 2
2 3 , 1z i z i = + = + . Tnh 21 2.z z ;
1 2z z v
1 23z z
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Dng Phc Sang - 49 - THPT Chu Vn An
Bi 13: Cho 2 3z i= + . Tm phn thc, phn o v mun ca 75
z i
iz
++
Bi 14: Cho ( )3311 2 2z i= + v ( )3
311 2 2
z i= + . Tnh1 2.z z
Bi 15: Gii cc phng trnh sau trn tp s phc:a) 2 2 0z + = b) 24 9 0z + = c) 2 4 8 0z z + =
d) 22 2 5 0z z+ + = e) 2 2 17 0z z+ + = f) 2 3 3 0z z + =
g) 3 4 0z z+ = h) 3 27 4z z z+ = i) 3 8 0z + =
j) 4 22 3 0z z+ = k) 4 22 3 5 0z z+ = l) 49 16 0z =
m) 22 4 9 0z z+ + = n) 2 1 0z z + = o) 2 4 11 0z z+ =
Bi 16: Tm s phc zc phn thc v phn o i nhau v 2 2z = Bi 17: Cho
1 2,z z l hai nghim phc ca phng trnh 25 2 1 0z z + =
Chng minh rng tng nghch o ca1
z v2
z bng 2.
Bi 18: Cho1 2,z z l hai nghim phc ca phng trnh 23 2 4 0z z + =
Chng minh rng1 2 1 2
. 2z z z z + + =
Bi 19: Cho1 2,z z l hai nghim phc ca phng trnh 2 4 5 0z z + =
Chng minh rng
2 2
1 2z 6z+ = Bi 20: Cho1 2,z z l hai nghim phc ca phng trnh 25 2 2 0z z + =
Chng minh rng1 2 1 2
.z z z z + =
Bi 21: Cho1 2,z z l hai nghim phc ca phng trnh 23 2 1 0z z + =
v2
z c phn o l mt s m. Tnh1 2
2z z+
Bi 22: Tm s phc z c phn thc v phn o bng nhau v 2 2z = Bi 23: Cho hai s phc ( 1)z m m i = + v 2 (2 3 )z n n i = + , vi
,m n . Tm z v z bit rng 1 7z z i+ = + .
Bi 24: Cho s phc ( 1) ,z m m i m = + + . Tm zbit rng 5z = .Bi 25: Cho s phc ( 1) ( 1) ,z m m i m = + + . Tm zbit . 10z z = .Bi 26: Cho s phc 2 ( 2) ,z m m i m = + + . Tm z bit rng 2z l
mt s phc c phn thc bng 5 .Bi 27: Gii cc phng trnh sau y trn tp cc s phc
a)5( 1)( 1) 2(4 5) 0z z z + + + = b) 22(2 1) (17 6) 0z z z + + =
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Ti liu tham kho - 50 - n tp tt nghip mn Ton
PPPhhhnnn VVV... PPPHHHNNNGGG PPPHHHPPP TTTOOO+ ++ TTTRRROOONNNGGG KKKHHHNNNGGG GGGIIIAAANNN
1. H to OxyzGm 3 trc Ox,Oy,Ozi mt vung gc nhau
c vct n v ln lt l: , ,i j k 2. To ca ima) nh ngha
( ; ; ) . . .M M M M M M
M x y z OM x i y j z k = + +
b) To ca cc im c bit Trung im Ica on AB
Trng tm Gca tam gic ABC
2
2
2
A BI
A BI
A BI
x x
x
y yy
z zz
+= +
= + =
3
3
3
A B CG
A B CG
A B CG
x x x
x
y y yy
z z zz
+ += + +
= + + =
Hnh chiu vung gc ca im ( ; ; )M M M
M x y z ln:
Trc Ox:1
( ;0;0)M
M x mp( )Oxy :12
( ; ; 0)M M
M x y
Trc Oy:2(0; ;0)
MM y mp( )Oxz :
13( ; 0; )
M MM x z
Trc Oz:3(0;0; )
MM z mp( )Oyz :
23(0; ; )
M MM y z
3. To ca vcta) nh ngha:
1 2 3 1 2 3( ; ; ) . . .a a a a a a i a j a k = = + +
b) Cng thc to ca vct
Nu ( ; ; ), ( ; ; )A A A B B B
A x y z B x y z th ( ; ; )B A B A B A
AB x x y y z z =
Nu1 2 3
( ; ; )a a a a = ,1 2 3
( ; ; )b b b b= th
1 1 2 2 3 3
( ; ; )a b a b a b a b+ = + + +
1 1 2 2 3 3
( ; ; )a b a b a b a b =
1 2 3. ( ; ; )k a ka ka ka =
, k c) iu kin cng phng ca hai vct
Cho 1 2 3( ; ; )a a a a =
, 1 2 3( ; ; )b b b b=
v 0b
. Khi ,a
cng phng vi b
tn ti s thc t sao cho .a t b=
1 1
2 2
3 3
a b
a b a b
a b
== = =
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Dng Phc Sang - 51 - THPT Chu Vn An
4. Tch v hng ca hai vct
a) Cng thc: Nu 1 2 3
1 2 3
( ; ; )
( ; ; )
a a a a
b b b b
= =
th1 1 2 2 3 3
. . . .a b a b a b a b= +
+
b) ng dng: 2 2 21 2 3
a a a a = + +
AB AB =
.
cos( , ).
a ba b
a b=
. 0a b a b =
, vi
0
0
a
b
5. Tch c hng ca hai vcta) nh ngha
Cho1 2 3
1 2 3
( ; ; )
( ; ; )
a a a a
b b b b
= =
. Khi , vct [ ]2 3 1 3 1 2
2 3 1 3 1 2, ; ;
a a a a a a
a b b b b b b b
=
c gi l tch c hng ca hai vct a
v b
.
b) Lu :Nu [ , ]n a b=
th n a
v n b
(gi s 0, 0, 0a b n
)
c) ng dng 1: Cho ba vct khc 0
ln lt l , ,a b c
. Khi ,
a
v b
cng phng vi nhau [ , ] 0a b =
,a b
v cng phng vi nhau [ , ]. 0a b c =
A,B,C thng hng [ , ] 0AB BC =
A,B,C,D ng phng [ , ]. 0AB AC AD =
d) ng dng 2: (tnh din tch)
Din tch hnh bnh hnh ABCD
[ , ]ABCD
S AB AD =
Din tch tam gic ABC:
ABCS
12
[ , ]AB AC=
e) ng dng 3: (tnh th tch) Th tch khi hnh hp .ABCD A B C D
[ , ].hh
V AB AD AA=
Th tch khi t din ABCD:
ABCDV = 16
[ , ].AB AC AD
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Ti liu tham kho - 52 - n tp tt nghip mn Ton
V D MINH HO
Bi 1: Trong h to ( , , , )O i j k cho 2 3OA i j k = + ,4 3 2 , (2; 7;1)OB i j k BC = + =
v (4;1; 7)A
a) Chng minh rng A,B,Cl 3 nh ca mt tam gic vung.b) Chng minh rng ( )AA ABC
c) Tnh th tch khi t din A ABC .d) Xc nh to cc nh cn li ca hnh hp .ABCD A B C D
Bi giiT gi thit ta c (2;1; 3), (4; 3; 2), (6; 4; 1), (4;1; 7)A B C A
Cu a:(2;2;1)
. 8 10 2 0(4; 5;2)
AB
AB AC AB AC AC
= = + = =
Vy, ABCl tam gic vung ti A
Cu b: Ta c, (2; 0; 4)AA =
v (2;2;1), (4; 5;2)AB AC = =
Do ,. 2.2 0.2 4.1 0
. 2.4 0.( 5) 4.2 0
AA AB
AA AC
= + = = + =
( )AA AB
AA ABC AA AC
Cu c:2 2 2
2 2 2
2 2 1 3
4 ( 5) 2 3 5
AB
AC
= + + = = + + =
. 9 5
2 2ABCAB AC
S = =
2 2 22 0 ( 4) 2 5h AA= = + + =
Vy, 9 5.2 51 13 3 3.2
. 15A ABC ABC
V h S AA = = = =B.
Cu d:ABCDl hnh bnh hnh AD BC =
2 2 4
1 7 6. (4; 6; 2)
3 1 2
D D
D D
D D
x x
y y D
z z
= = = = + = =
Tng t, (6;3; 6)B , (6; 6; 6)D , (8; 4; 5)C
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BI TP V TO CA IM, TO CA VCTBi 2: Trong h to Oxyz, cho cc im (2; 0; 1), (3;2; 3), ( 1;1;1)A B C
a) Chng minh rng A,B,Cl ba nh ca mt tam gic.b) Xc nh to nh Dv tm Ica hnh bnh hnh ABCD.
c) Tm to im Msao cho 2AM OB AC
=
Bi 3: Trong h to Oxyz, cho cc im (2;2; 1), (2;1; 0), (1;1; 1)A B C
a) Chng minh rng ABCl tam gic u.b) Cho im (4;0; 3)A . Xc nh to cc im B v C
.ABC A B C l mt hnh lng tr.c) Chng minh rng .ABC A B C l mt lng tr u.
Bi 4: Trong h to ( , , , )O i j k cho 3 2 3OM i j k = + v A,B,Clnlt l hnh chiu vung gc ca Mln cc trc to Ox,Oy,Oz.a) Chng minh rng ABCl tam gic cn.b) Tnh th tch t din OABC, t tnh khong cch t gc to n mt phng ( )ABC
Bi 5: Trong h to ( , , , )O i j k cho 3 2 3ON i j k = + v A,B,Clnlt l hnh chiu vung gc ca im Nln cc mt phng to Oxy, Oyz, Oxz.
a) Tnh din tch tam gic ABCv th tch ca t din NABC.b) Tnh khong cch t im Nn mt phng ( )ABC Bi 6: Trong khng gian vi h to Oxyz, chng minh rng (0;0;0)O ,
A(0;1;2),B(2;3;1),C(2;2;1) l bn nh ca mt hnh ch nht.
Bi 7: Trong h to ( , , , )O i j k cho t din ABCD sao cho(2; 4; 1), 4 , (2; 4; 3), (0; 2; 0)A OB i j k C AD = + =
a) Chng minh rng AB, ACv ADi mt vung gc vi nhau.
b) Tnh din tch tam gic ABCv th tch t din ABCD.Bi 8: Trong h to Oxyzcho (2;1; 3), (4; 3; 2), (6; 4; 1)A B C
a) Chng minh rng A,B,Cl ba nh ca mt tam gic vung.b) Tm to im D A,B,C,Dl 4 nh ca mt hnh ch nht
Bi 9: Tm to cc nh cn li ca hnh hp .ABCD A B C D bitrng (2;4; 1), (1;4; 1), (2;4;3), (2;2; 1)A B C OA
=
Bi 10:Tm im Ntrn Oycch u hai im (3;1;0)A v ( 2; 4;1)B Bi 11:Tm im M trn mt phng ( )Oxz cch u ba im (1;1;1)A ,
( 1;1; 0)B v (3;1; 1)C
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Ti liu tham kho - 54 - n tp tt nghip mn Ton
6. Phng trnh mt cua) Dng 1: mt cu ( )S tm I(a;b;c), bn knh R c phng trnh:
( ) ( )2 2 2 2( ) x a y b z c R+ + =
b) Dng 2: vi iu kin2 2 2
0a b c d + + > th2 2 2 2 2 2 0x y z ax by cz d + + + = l phng trnh mt cu Tm I(a;b;c)
Bn knh 2 2 2R a b c d = + + c) Lu : mt cu ( , )S I R tip xc vi mt phng ( ) ( , )d I R = 7. Phng trnh tng qut ca mt phnga) Cng thc: Nu mt phng ( )P i qua im
0 0 0 0
( ; ; )M x y z v c vct
php tuyn ( ; ; ) 0n = A B C th ( )P c phng trnh tng qut l:
0 0 0( ) ( ) ( ) 0A x x B y y C z z + + =
b) Lu v cch xc nh vct php tuyn (vtpt) cho mt phng:
Nu ( )P AB th ( )P nhn n AB=
lm vct php tuyn.
Nu a
v b
l hai vct khng cng phng, c gi song song hoc
cha trong ( )P th ( )P nhn [ , ]n a b
=
lm vct php tuyn.
Cho trc ( ) : 0Q Ax By Cz D + + + = . Nu ( )( )P Q th ( )P c
phng trnh dng 0Ax By Cz D + + + = (vi D D )
Mt phng ( ) : 0P Ax By Cz D + + + = c vtpt ( ; ; )n A B C = c) Phng trnh mt phng theo on chn
Mt phng ( )P i qua ba im phn bit
( ;0;0)A a , (0; ;0), (0; 0; )B b C c c phng trnh
1x y z
a b c+ + =
d) Khong cch t im Mo n mt phng (P)
0 0 0
2 2 20( ,( )) Ax By Cz D
A B C
d M P + + ++ +
=
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Dng Phc Sang - 55 - THPT Chu Vn An
8. Phng trnh ca ng thngCho ng thng di qua im
0 0 0 0( ; ; )M x y z v c vtcp ( ; ; )u a b c =
a) Phng trnh tham s ca d:
0
0
0
( )
x x at
y y bt t z z ct
= += + = +
b) Phng trnh chnh tc ca d: 0 0 0x x y y z z
a b c
= =
(gi s a,b,cu khc 0)c) Cch xc nh vct ch phng (vtcp) cho ng thng d
di qua 2 im A v Bphn bit th dc vtcp u AB=
Cho ng thng c vtcp u . Nu d th dc vtcp u u=
Cho mt phng ( )P c vtptP
n
. Nu d(P) th dc vtcpP
u n=
Cho hai vct khng cng phng a
v b
. Nu dvung gc vi gi
ca 2 vct a
v b
th dc vtcp [ , ]u a b
=
Cho ng thng c vtcp u
v mt phng ( )P c vtptP
n
. Nu
dsong song vi ( )P v vung gc vi th dc vtcp [ ],P
u n u=
Cho hai mt phng ( )P v ( )Q ln lt c vtptP
n
vQ
n
.
Nu dl giao tuyn ca ( )P v ( )Q th dc vtcp [ ],P Q
u n n=
Cho hai ng thng1
d v2
d ln lt c vtcp1
u
v2
u
khng
cng phng. Nu dvung gc vi c hai ng thng1
d v2
d th
dc vtcp1 2
[ , ]u u u=
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Ti liu tham kho - 56 - n tp tt nghip mn Ton
V D MINH HOBi 12: Cho A(1;3;1), B(2;1;2), C(0;2; 6) v ( ) : 2 2 1 0P x y z + + =
a) Vit phng trnh mt cu tm B, i qua Ab) Vit phng trnh mt cu ng knh BC.
c) Vit phng trnh mt cu tm C, tip xc vi mt phng ( )P d) Vit phng trnh mt cu ngoi tip t din OABC.
Bi giiCu a:Gi
1( )S l mt cu tm B(2;1;2)v i qua im A. Khi
1( )S
c bn knh1
R AB=
Ta c 2 2 2(1; 2;1) 1 ( 2) 1 6AB AB = = + + =
1
( )S c phng trnh 2 2 2( 2) ( 1) ( 2) 6x y z + + =
Cu b: Gi2
( )S l mt cu ng knh BCth2
( )S c tm (1 2)32
; ;I l
trung im ca on thng BCv bn knh2
BCR =
v 2 2 2( 2;1; 8) ( 2) 1 ( 8) 69BC BC = = + +
=
nn 692 2
BCR = =
Phng trnh mt cu 2( )S l2 2 23 69
2 4( 1) ( ) ( 2)x y z + + + = Cu c: Gi
3( )S l mt cu tm C(0;2;6), tip xc vi ( )P . Khi
3( )S
c bn knh3