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8/10/2019 Bai Tap Toan Lop 11(Full)
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Trng THPT Ng Thi Nhim Bi tp ton 11
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Trng THPT Ng Thi Nhim Bi tp ton 11
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2
Bi 1: Tm
a)6
293lim
3
23
2
+ xx
xxx
x b)
21
3 2lim
1x
x
x
+
Bi 2: Xt tnh lin tc ca hm ssau trn tp xc nh ca n: + +
= +
23 2
, khi x 2( ) 2
3 , khi x = -2
x x
f x x
Bi 3: Cho hm sy = f(x) = 2x3 6x +1 (1)a) Tm o hm cp hai ca hm s(1) ri suy ra ( 5)f .b)
Vit phng trnh tip tuyn ca th hm s (1) tiim Mo(0; 1).
c) Chng minh PT f(x) = 0 c t nht mt nghim nmtrong khong (-1; 1).
Bi 4: Cho hnh chp S. ABCD c y ABCD l hnh thoi cnh
a c gc BAD = 600
v SA=SB = SD = a.a)
Chng minh (SAC) vung gc vi (ABCD).b) Chng minh tam gic SAC vung.c) Tnh khong cch tS n (ABCD).
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MT STHI THAM KHO
1
Cu 1: Tnh gii hn ca hm s
a)
2
3
2 9 9lim
3xx x
x
b)
22 4 1lim
3 2xx x
x
+
+
Cu 2: Xt tnh lin tc ca hm strn tp xc nh ca n:
f(x) =
22 10 22 4
4 17 2
x xx
x
x x
+ + < +
+
nu
nu
Cu 3: Tnh o hm ca cc hm s:a) y = 3x3- 4x2+ 8
b) y =22 5 1
3 4
x x
x
+
c) y = 3sin3x - 3cos24xCu 4:
a) Vit phng trnh tip tuyn ca thhm s(C)y = - 2x4+ x2 3 ti im thuc (C) c honh x0= 1.
b) Cho hm sy = x.cosx.Chng minh rng: x.y 2(y - cosx) + x.y = 0
Cu 5: Cho hnh chp S.ABC c y l tam gic cn B vABC=1200, SA (ABC) v SA = AB = 2a. Gi O l trungim ca on AC, H l hnh chiu ca O trn SC.
a) Chng minh: OB SC.b) Chng minh: (HBO) (SBC).c) Gi D l im i xng vi B qua O. Tnh khong
cch gia hai ng thng AD v SB.
Trng THPT Ng Thi Nhim Bi tp ton 11
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Chng I:HM SLNG GIC PHNG TRNH
LNG GIC
PHN 1. HM SLNG GIC
Bi 1.Tm tp xc nh ca cc hm ssau:
1.1
sin1
+=
xy
x 2.
3sin2
2cos3=
xy
x
3. cot(2 )4
= y x 4. 2tan( 5 )
3= +y x
5.1
cos1
=
+
xy
x 6.
sin 2
cos 1
+=
+
xy
7.1
sin cos=
y
x x 8.
2 2
3 tan
cos sin
+=
xy
x x
9. sin coscos 1 1 sin
= + +
x xyx x
10. 212 sin tan 1= + y x
x
Bi 2.Xc nh tnh chn, lca cc hm s:
1.cos3x
yx
= 2. 2 2siny x x=
3. 2siny x x= + 4. 21
tan 12
y x= +
5. 23sin cosy x x= 6. tan 2 cosy x x= + Bi 3.Tm gi trln nht, gi trnhnht ca cc hm s:
1. y 2sin(x ) 33
= + 2.1
y=3- cos2x2
3.21 3cos
y=2
x+ 4. 2 4sin cosy x x=
5.2
4sin cos2y x x= 6. 3 cos2 1y x= +
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7. 7 3 sin3y x= 8. 2 25 2sin cosy x x=
Bi 4.Hy xt sbin thin v vthcc hm ssau:1. siny x= 2. 2 siny x=
3. sin( )3
y x
= + 4. cos 1y x= +
PHN 2. PHNG TRNH LNG GIC
DNG 1. PHNG TRNH LNG GIC CBN
Bi 1. Gii cc phng trnh sau:
1.1
sin32
x= 2.2
cos22
x=
3. tan( ) 34
x
= 4. s in2 s in2 cos 0x x x =
5. s in3 cos2 0x x = 6. t an4 cot 2 1x x=
7. 2 cos( ) 1 06
x
+ = 8. tan(2 ) t an3 03
x x
+ + =
9. 2cos 2sin 02
xx = 10. 4 4
2cos sin
2x x =
11.1
sin cos sin cos2 3 3 2 2
x x + =
12. 3 32
sin cos cos sin8
x x x x =
13. 2 2 2cos cos 2 cos 3 1x x x+ + =
14.2 2 17sin 2 cos 8 sin( 10 )
2x x x
= +
15.
4 6
cos sin cos2x x x+ =
Trng THPT Ng Thi Nhim Bi tp ton 11
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3. Dng v tnh di on vung gc chung ca AB v
SD
4.
Tnh : d[ ])(, SACM
Bi 6. Cho hnh lng trABC.ABCc AA(ABC) v AA
= a, y ABC l tam gic vung ti A c BC = 2a, AB = a 3 .1. Tnh khong cch tAAn mt phng (BCCB).2. Tnh khong cch tA n (ABC).3. Chng minh rng AB (ACCA) v tnh khong cch
tAn mt phng (ABC).
Bi 7. Cho hnh lp phng ABCD.ABCD.
1.
Chng minh: BD (BAC); BD (ACD)
2. Tnh d (BA'C'),(ACD')
3.
Tnh d (BC'),(CD')
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1. OA v BC 2. AI v OC.Bi 2. Cho hnh chp SABCD, y ABCD l hnh vung tm O,cnh a, SA (ABCD) v SA = a. Tnh khong cch gia haing thng:
1. SC v BD. 2. AC v SD.
Bi 3. Cho hnh chp S.ABCD c y ABCD l hnh vung
canh a, SA (ABCD) v SA = 3a . Tnh:
1. Gia SC v BD ; gia AC v SD.
2.
d [ ])(, ABCDA
3.
d [ ])(, SBCO vi O l tm ca hnh vung.
4.
d [ ])(, ABCDI vi I l trung im ca SC.
Bi 4. Cho hnh chp S.ABCD c y ABCD l hnh thang
vung ti A v D AB = DC = a , SA (ABCD) v SA = 2a
Tnh :
1. d [ ])(, SCDA ; d [ ])(, SBCA
2. d [ ])(, SCDAB
3. d [ ])(, SCDAB
4.
d [ ])(, SBCDE , E l trung im ca ABBi 5. Cho hnh chp S.ABCD c y l hnh vung cnh a ,tam
giac SAD u v (SAD) (ABCD) .gi I l trung im ca Sb
va K =CM BI
1. Chng minh (CMF) (SIB)
2.
Chng minh : tam giac BKF cn ti K
Trng THPT Ng Thi Nhim Bi tp ton 11
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16.1 cos4 s in4
02s in2 1 cos4
x x
x x
=
+
17. 2 2 1sin cos cos2
x x x ++ =
18.
2(2 3)cos 2sin ( )2 4 1
2 cos 1
xx
x
=
Bi 2.Gii v bin lun phng trnh:1. sin 2 1x m=
2. (4 1)cos cos 8m x m x = 3. 4tan ( 1) tanx m m x = +
4. 2(3 2)cos2 4 sin 0m x m x m + + = Bi 3.Tm m phng trnh:
1. 2 sin( )4
x m
+ = c nghim (0; )2
x
2.7
(2 )sin( ) (3 2)cos(2 ) 2 02m x m x m
+ + + + = cnghim.
DNG 2. PHNG TRNH BC HAI I VI MTHM SLNG GIC
Bi 1.Gii cc phng trnh sau:
1. 24cos 2( 3 1)cos 3 0x x + + = 2. 22cos x 5sinx 4 0+ = 3. 2cos2x 8cosx 5 0+ = 4. 2cosx.cos2x 1 cos2x cos3x= + +
5. 22
33 2 tan
cos = + xx
6. 5tan x 2cotx 3 0 =
7. 26sin 3 cos12 4x x+ =
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8. 2cos2 3cos 4 cos2
x xx
=
9. 2cos4cot tansin2
xx xx
= +
10.2cos (2sin 3 2) 2sin 3
11 sin2
x x x
x
+ + =
+
11. 4 43tan 2 tan 1 0x x+ =
12.1 1
cos sinsin cos
x xx x
=
13. 22
1 1cos 2(cos ) 1
coscosx x
xx+ + =
14.2 2
1 14
sin cossin cos x xx x+ =
Bi 2.Tm m phng trnh sau c nghim:
1. 2cos (1 )cos 2 6 0x m x m+ + =
2. 24 cos 2 4 cos2 3 3 0x x m =
Bi 3.Cho phng trnh: cos2 ( 2)sin 1 0x a x a+ + = 1. Gii phng trnh cho khi a = 1.
2. Vi gi trno ca a th phng trnh cho cnghim?
DNG 3. PHNG TRNH BC NHT THEOSINu V COSu
Bi 1.Gii cc phng trnh sau:
1. 2sincos3 = xx
2. 1sin3cos = xx
Trng THPT Ng Thi Nhim Bi tp ton 11
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1. Chng minh: (SAB) (SAD), (SAB) (SBC).
2. Tnh gc gia hai mp (SAD), (SBC).
3.
Gi H, I ln lt l trung im ca AB v BC. Chng
minh: (SHC) (SDI).
Bi 10. Cho tam gic ABC vung ti A. Gi O, I, J ln lt l
trung im ca BC v AB, AC. T O k on thng
OS (ABC).
1.
Chng minh: (SBC) (ABC).2. Chng minh: (SOI) (SAB).
3. Chng minh: (SOI) (SOJ).
Bi 11. Cho tam din ba gc vung Oxyz (3 tia Ox, Oy, Oz i
mt vung gc). Ln lt ly trn Ox, Oy, Oz cc im B, C, A
sao cho OA = a, OB = b, OC = c. Cc ng cao CH va BK catam gic ABC ct nhau ti I.
1. Chng minh: (ABC) (OHC).
2. Chng minh: (ABC) (OKB).
3. Chng minh: OI (ABC).
4. Gi , , ln lt l gc to bi OA, OB, OC vi OI.
Chng minh: cos2+ cos2+ cos2= 1.
KHONG CCH
Bi 1. Cho hnh tdin OABC, trong OA, OB, OC = a. Gi I
l trung im ca BC. Hy dng v tnh di on vung gcchung ca cc cp ng thng:
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1. Chng minh: (SBC) (ABC).
2. Chng minh: (SOI) (ABC).
Bi 6. Cho hnh chp S.ABCD, y l hnh vung cnh a. Tam
gic SAB u nm trong mt phng vung gc vi y. I, J, K
ln lt l trung im ca AB, CD, BC.
1. Chng minh: SI (ABCD).
2. Chng minh: trn mt phng SAD v SBC l nhng tam
gic vung.3.
Chng minh: (SAD) (SAB), (SBC) (SAB).
4.
Chng minh: (SDK) (SIC).
Bi 7. Cho tdin ABCD c cnh AD (BCD). Gi AE, BF
l hai ng cao ca tam gic ABC, H v K ln lt l trc tm
ca tam gic ABC v tam gic BCD.1. Chng minh: (ADE) (ABC).
2.
Chng minh: (BFK) (ABC).
3.
Chng minh: HK (ABC).
Bi 8.Trong mp (P) cho hnh thoi ABCD vi AB = a, AC =
2 63
a . Trn ng thng vung gc vi mp (P) ti giao im O
ca hai ng cho hnh thoi ta ly S sao cho SB = a.
1. Chng minh: SAC vung.
2.
Chng minh: (SAB) (SAD).
Bi 9. Cho hnh vung ABCD. Gi S l im trong khng gian
sao cho SAB l tam gic u v (SAB) (ABCD).
Trng THPT Ng Thi Nhim Bi tp ton 11
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3. s in3 3 cos3 2x x+ =
4. 22 cos 3 s in2 2x x =
5. 2s in2 cos2 3 cos 4 2 0x x x+ + =
6. )7sin5(cos35sin7cos xxxx =
7.4
1)
4(cossin 44 =++
xx
8. tan 3cot 4(sin 3 cos )x x x x = +
9.2 1
sin 2 sin 2x x+ =
10. 33sin3 3 cos9 1 4sin 3x x x = +
11.3(1 cos 2 )
cos2sin
xx
x
=
12.cos sin
cot tansin cos
x xx x
x x
=
Bi 2. nh m phng trnh sau y c nghim:1. sin 2 cos 3m x x+ = 2. s in2 cos2 2 0x m x m+ + = 3. cos3 ( 2)s in3 2m x m x+ + = 4. (sin 2cos 3) 1 cosx x m x+ + = + 5. (cos sin 1) sinm x x x = 6. (3 4 )cos2 (4 3)s in2 13 0m x m x m+ + + =
Bi 3. Cho phng trnh: sin cos 1x m x+ = 1. Gii phng trnh khi 3m= .2. nh m phng trnh trn v nghim.
DNG 4. PHNG TRNH THUN NHT BC HAITHEO SINu V COSu
Bi 1.Gii cc phng trnh sau:1. 2 2sin x 3sinxcosx 4cos x 0+ =
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2. 2 23sin x 8sinxcosx ( 8 3 9)cos x 0+ + =
3. 2 24sin x 3
sin2x 2cos x 4+ =
4. 2 22sin x 5sinx.cosx cos x 2=
5. 2 24sin 3 3 sin 2 cos 42 2
x xx+ =
6. 2 22sin 6sin cos 2(1 3)cos 5 3x x x x+ + + = +
7. 3 2 3sin 2sin cos 3cos 0x x x x+ =
8. 3 2 34sin 3sin cos sin cos 0x x x x x+ =
9. 3 3 2 2sin 3 cos sin cos 3 sin cosx x x x x x =
10.2
2 tan cot 3sin2
x xx
+ = +
Bi 2.Tm m phng trnh sau c nghim:
1. 2 2sin 2s in2 3 cos 2m x x m x+ + =
2. 2 2sin s in2 ( 1)cos 0x m x m x + =
DNG 5. PHNG TRNH I XNG PHN XNG
Bi 1.Gii cc phng trnh sau:1. 2(sin cos ) 3sin cos 2 0x x x x+ + + =
2. ( )3 sinx cosx 2sin2x 3 0+ + + =
3. ( )sin2x 12 sinx cosx 12=
4. ( )2 cosx sinx 4sinxcosx 1+ = + 5. cosx sinx 2sin2x 1 0=
6. (1 2)(sin cos ) 2sin cos 1 2 0x x x x+ + =
7. 3 3sin cos 1 sin cosx x x x+ =
8. 3 3sin cos 2(sin cos ) 1x x x x+ = +
9. tan cot 2(sin cos )x x x x+ = +
Trng THPT Ng Thi Nhim Bi tp ton 11
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3. Gi BE, DF l hai ng cao ca tam gic SBD. Chng
minh rng: (ACF) (SBC), (AEF) (SAC).
Bi 2. Cho tdin ABCD c cc mt ABD v ACD cng vung
gc vi mt BCD. Gi DE ,BK l ng cao tam gic BCD v
BF l ng cao tam gic ABC
1. Chng minh : AD (BCD)
2. Chng minh : (ADE) (ABC)
3.
Chng minh : (BKF) (ABC)4. Chng minh : (ACD) (BKF)
5. Gi O v H ln lt l trc tm ca hai tam gic BCD v
ABC chng minh : OH (ABC)
Bi 3. Cho hnh chp S.ABCD c y ABCD l hnh thoi cnh
a. SA= SB= SC=a. Chng minh :1. (ABCD) (SBD)
2. Tam gic SBD l tam gic vung.
Bi 4. Cho tam gic u ABC cnh a, I l trung im ca cnh
BC, D l im i xng ca A qua I. Dng on SD =6
2
a
vung gc vi (ABC). Chng minh:
1. (SAB) (SAC).
2.
(SBC) (SAD).
Bi 5. Cho hnh chp S.ABCD c y ABC l tam gic l tam
gic vung ti A, AB = 2a, AC = a, SA = SB = SC = 2a . Gi
O l trung im ca BC, I l trung im ca AB.
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3. Tnh gc [(SMC), (ABC)].
Bi 7. Cho hnh chp S.ABCD c y ABCD l hnh thang
vung ti A v D vi AB = 2a, AD = DC = a, SA = 2a . SA
(ABCD). Tnh gc gia cc mt phng.
1. (SBC) v (ABC).
2.
(SAB) v (SCB).
3. (SCB) v (SCD).
Bi 8. Cho hnh chp S.ABCD c y l hnh thoi ABCD tm
O, cnh a ABC= 600, SO (ABCD) v SO =3
4
a. Tnh so
nhdin cnh AB.
Bi 9. Cho hnh chp S.ABCD c y ABCD l hnh vung
cnh a, tm O, SA
(ABCD) v SA = x (x>0).1.
Tnh s[S, BC, A] theo a v x. Tnh x theo a so nh
din trn bng 600.
2.
Tnh s[B, BC, D] theo a v x. Tnh x theo a so nh
din trn bng 1200
HAI MT PHNG VUNG GC
Bi 1. Cho hnh chp S.ABCD, y ABCD l hnh vung, SA
(ABCD).
1.
Chng minh: (SAC) (SBD).
2. Chng minh: (SAD) (SCD), (SAB) (SBC).
Trng THPT Ng Thi Nhim Bi tp ton 11
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10.cos2
sin cos1 s in2
xx x
x+ =
Bi 2.nh m phng trnh sau c nghim:1. sin cos 1 s in2x x m x+ = +
2. 2s in2 2 2 (sin cos ) 1 6 0x m x x m + + =
DNG 6. PHNG TRNH LNG GIC KHNG MUMC
Bi tp. Gii cc phng trnh sau:1. sin .s in2 1x x=
2. 2 1007cos 8sin 8x x+ =
3. sin cos 2(2 s in3 )x x x+ =
4. 3 3 4sin cos 2 s inx x x+ =
MT STHI I HC1. 2(1 2sin ) cos 1 sin cosx x x x+ = + +
2. 3 cos5 2sin 3 cos 2 sin 0x x x x =
3. 3sin cos sin 2 3 cos3 2(cos 4 sin )x x x x x x+ + = +
4.(1 2sin ) osx
3
(1 2sin )(1 s inx)
x c
x
=
+
5. sin 3 3 cos3 2sin 2x x x = 6. 2sin (1 cos 2 ) sin 2 1 2cosx x x x+ + = +
7. 3 3 2 2sin 3 cos sin cos 3 sin cosx x x x x x =
8.1 1 7
4sin( )3sin 4sin( )2
xx
x
+ =
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9. 2(sin cos ) 3 cos 22 2
x xx+ + =
10.
2
2sin 2 sin 7 1 sinx x x+ =
11. 2 2(1 sin )cos (1 cos )sin 1 sin 2x x x x x+ + + = + 12. cos3 cos 2 cos 1 0x x x+ =
13. cot sin (1 tan tan ) 42
xx x x+ + =
14.6 62(cos sin ) sin cos
02 2sin
x x x x
x
+ =
15. 4 4 3cos sin cos( )sin(3 ) 04 4 2
+ + =x x x x
16. 1 sin cos sin 2 cos 2 0x x x x+ + + + = 17. 2 2cos 3 cos 2 cos 0x x x = 18. 25sin 2 3(1 sin ) tanx x x = 19. (2cos 1)(2sin cos ) sin 2 sinx x x x x + =
20.
2
cot tan 4sin 2 sin2x x x x + =
Trng THPT Ng Thi Nhim Bi tp ton 11
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Bi 4. Cho hnh vung ABCD v tam gic u SAB cnh a nm
trong hai mt phng vung gc nhau. Gi I l trung im ca
AB.
1. Chng minh: SI (ABCD) v tnh gc gia SC v
(ABCD).
2. Gi J l trung im CD. Chng t: (SIJ) (ABCD) . Tnh
gc hp bi SI v (SDC).
Bi 5. Cho hnh chp S.ABCD c y ABCD l hnh vung tm
O, cnh a, SA (ABCD) v SA = a. Tnh:
1.
[SAB, (SCD)].
2. [SAB, (SBC)].
3. [SAB, (SAC)].
4.
[SCD, (ABCD)].5. [SBC, (SCD)].
6. s[S, BC, A].
7. s[C, SA, D].
8.
s[A, SB, D].
9.
s[B, SC, A].Bi 6. Cho hnh chp S.ABCD c y ABC l tam gic vung
ti B, AB = 2a, BC = 3a , SA (ABC) v SA = 2a. Gi M l
trung im ca AB.
1. Tnh gc [(SBC), (ABC)].
2.
Tnh ng cao AK ca AMC.
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4. Gi d l ng thng vung gc vi (ABC) ti trung im
K ca BC tm d ( ).
- GC GIA NG THNG V MT PHNG
- GC GIA HAI MT PHNG
Bi 1. Cho hnh chp S.ABCD c y ABCD l hnh vung
cnh a, tm O, SO (ABCD), M, N ln lt l trung im ca
SA v BC, bit 0( ,( )) 60MN ABCD = .
1. Tnh MN v SO.
2. Tnh gc gia MN v mp(BCD).
Bi 2. Cho hnh chp S.ABCD c y ABCD l hnh vung
cnh a. SA (ABCD) v SA = a 6 . Tnh gc gia:
1. SC v (ABCD)
2.
SC v (SAB)
3. SC v (SBD)
4.
SB v (SAC)
Bi 3. Cho t din ABCD c AB (BCD) v AB = 3a ,
BCD l tam gic u cnh a. Tnh gc gia:
1. AC v (BCD).
2. AD v (BCD).
3. AD v (ABC).
Trng THPT Ng Thi Nhim Bi tp ton 11
11
Chng II. T HP XC SUT
PHN 1. HON VN- CHNH HP - THP
Bi 1.C 25 i bng tham gia thi u, c2 i th vi nhau2 trn ( i v v). Hi c tt cbao nhiu trn u?Bi 2.
1. Tcc chs1, 2, 3, 4, 5 c thlp c bao nhiu stnhin c 5 chs?
2.
Tcc chs0, 1, 2, 3, 4, 5, 6 c thlp c baonhiu stnhin c 3 chsv l schn?3. C bao nhiu stnhin c 6 chsi mt khc nhauv chia ht cho 5?
Bi 3.Mt hi ng nhn dn c 15 ngi, cn bu ra 1 chtch, 1 ph chtch, 1 thk. Hi c my cch nu khng aic kim nhim?Bi 4.Trong mt tun, An nh mi ti i thm 1 ngi bn
trong s10 ngi bn ca mnh. Hi An c thlp c baonhiu khoch thm bn nu:1.
C ththm 1 bn nhiu ln?2. Khng n thm 1 bn qu 1 ln?
Bi 5. C bao nhiu cch xp 10 hc sinh thnh mt hng dc?Bi 6. C bao nhiu cch xp 5 bn A, B,C,D,E vo mt ghdi5 chnu:
1. Bn C ngi chnh gia.
2.
Hai bn A v E ngi hai u gh.Bi 7.Tcc chs1,2,3,4,5,6 c ththit lp c bao nhiusc 6 chskhc nhau m hai chs1 v 6 khng ng cnhnhau?Bi 8.C 2 sch Ton khc nhau, 3 sch L khc nhau v 4sch Ha khc nhau.Cn sp xp cc sch thnh mt hng saocho cc sch cng mn knhau. Hi c bao nhiu cch?Bi 9.Gii :
1.
P2.x2 P3.x = 8
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2.
1
1
1
6
x x
x
P P
P
+
=
3.
12
4 15. +
+ ++
++
+ nnn Bi 4.Chng minh vi mi stnhin 2n , ta c cc btng thc sau:
1. 133 +> nn
2.2
32 > nn
3. 322 1 +>+ nn
Bi 5.Chng minh vi mi stnhin 3n , ta c:122 +> nn
DY S
Bi 1. Xt tnh n iu cc dy ssau :
1. 21 1nu
n= + 2. 12 3+= n
n
nu
3.1
2
n
nu =
4. nnun += 1 .
5.2 1
2
n
n nu
= 6.
nn
nu
2
2+=
7. nu nn
= 3 8. 12 = nnun
.
Bi 2.Xt tnh bchn cc dy ssau :
1. 23 = nun 2.1
( 1)nu
n n=
+
3. 13.2nnu = 4. nnu )3(=
5.34
34
+
=
n
nun 6.
2
1
1n
nu
n
=
+
Trng THPT N g Thi N him Bi tp ton 11
47
2. GisAB CD th MN QG l hnh g? Tnh SMN PQbitAM = x, AB = AC = CD = a. Tnh x din tch ny lnnht.
HAI MT PHNG SONG SONG
Bi 1. Cho hai hnh bnh hnh ABCD , ABEF c chung cnhAB v khng ng phng . I, J, K ln lt l trung im ca cccnh AB, CD, EF. Chng minh:
1. (ADF) // (BCE).
2.
(DIK) // (JBE).Bi 2.Cho tdin ABCD.Gi H, K, L l trng tm ca cc tamgic ABC, ABD, ACD. Chng minh rng (HKL)//(BCD).Bi 3. Cho hnh chp S.ABCD y l hnh bnh hnh tm O.Tam gic SBD l tam gic u. Mt mp () di ng song songvi (SBD) qua im I trn on AC. Xc nh thit din cahnh chp ct bi ().Bi 4.Cho hnh chp S.ABCD c ABCD l hnh thang vung
ti A v D; AD = CD = a ; AB = 2a, tam gic SAB vung cntiA.Trn cnh AD ly im M. t AM =x. Mt phng () quaM v //(SAB).
1.
Dng thit din ca hnh chp vi ().2. Tnh din tch v chu vi thit din theo a v x.
Bi 5.Cho hai mp (P) v (Q) song song vi nhau v ABCD lmt hnh bnh hnh nm trong mp (P). cc ng thng song
song i qua A, B, C, D ln lt ct mp (Q) ti cc im A', B',C', D'.1. Tgic A'B'C'D' l hnh g?2. Chng minh (AB'D') // (C'BD).3.
Chng minh rng on thng A'C i qua trng tm ca haitam gic AB'D' v C'BD. Hai mp (ABD), (CBD) chiaon A'C lm ba phn bng nhau.
HNH LNG TR
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Trng THPT N g Thi N him Bi tp ton 11
46
Chng minh : MN // (BCD) v MN // (ABC).Bi 2.Cho tdin ABCD .Gi I, J l trung im ca BC v CD
1.
Chng minh rng BD//(AIJ)
2.
Gi H, K l trng tm ca cc tam gic ABC v ACD.Chng minh rng HK//(ABD)
Bi 3.Cho hnh chp S.ABCD c ABCD l hnh bnh hnh .Gl trng tm ca tam gic SAB v E l im trn cnh AD saocho DE = 2EA. Chng minh rng GE // (SCD).Bi 4.Hnh chp S.ABCD c y ABCD l hnh bnh hnh.Gi M , N theo thtl trung im ca cc cnh AB, CD .
1. Chng minh MN // (SBC) v MN // (SAD)2.
Gi P l trung im ca cnh SA. Chng minh SB //(MN P) v SC // (MN P).
Bi 5.Cho hnh chp S.ABCD. M, N l hai im bt k trnSB v CD. () l mt phng qua MN v song song vi SC.
1. Tm cc giao tuyn ca ( ) vi cc mt phng (SBC),(SCD) v (SAC).
2.
Xc nh thit din ca S.ABCD vi mt phng () .
Bi 6.Cho hnh chp S.ABCD c ABCD l hnh bnh hnh .GiM,N l trung im SA,SB. im P thay i trn cnh BC
1. Chng minh rng CD//(MN P)2.
Dng thit din ca hnh chp vi mt phng (MN P) .Chng minh rng thit din l 1 hnh thang.
3. Gi I l giao im 2 cnh bn ca thit din ,tm qutchim I
Bi 7. Cho hnh chp S.ABCD. M, N l hai im trn AB,CD, () l mt phng qua MN v song song vi SA.
1. Xc nh thit din ca hnh chp v mt phng ().2. Tm iu kin ca MN thit din l hnh thang.
Bi 8. Cho tdin ABCD. Tim M trn AC ta dng mt mp() song song AB v CD. Mp ny ln lt ct BC, BD, AD tiN , P, Q.
1.
Tgic MN QG l hnh g?
Trng THPT N g Thi N him Bi tp ton 11
19
Bi 3. Cho dy s ( )nu xc nh bi:
+
+=
=
+
1
2
1
1
1
n
n
n
u
uu
u
; 1n .
Chng minh rng nu bchn trn bi 2
3v bchn di bi 1.
Bi 4.Cho dy s ( )nu xc nh bi:
+=
=
+ 2
1
2
1
1
n
n
uu
u
; 1n .
Chng minh rng nu l dy gim v bchn.
Bi 5.Cho dy s ( )nu xc nh bi:
++=
=
+n
nn nuu
u
2).1(
1
1
1
; 1n .Chng minh rng :
1. ( )nu l dy tng.
2. nn nu 2).1(1 += , 1n .
CP SCNG
Bi 1.Tm shng u v cng sai ca cc cp scng, bit :
1.
=+
=+
17
10
61
531
uu
uuu 2.
=
=
75
8
152
37
uu
uu
3.
=
=+
129
14
12
53
s
uu
4.
=+
=+
1170
60212
24
157
uu
uu
5.
=
=++
24
25
82
541
uu
uuu 6.
=
=
75.
8
72
37
uu
uu
Bi 2.1. Cho cp scng c 1a =10, d = -4 .Tnh 10a v 10S .
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Trng THPT N g Thi N him Bi tp ton 11
44
2. Tm giao im ca SD vi mt phng (AMN ) ?3.
Tm tit din to bi mt phng (AMN ) vi hnh chpBi 18: Hnh chp SABCD c y ABCD l hnh bnh hnh .
M l trung im SC1. Tm giao im I ca AM vi (SBD) ? Chng minh IA= 2IM .
2. Tm giao im F ca SD vi (AMB) ? Chng minh F ltrung im SD ?
3.
Xc nh hnh dng tit din to bi (AMB) vi hnhchp.
4. Gi N l mt im trn cnh AB .Tm giao im ca
MN vi (SBD) ?
HAI NG THNG SONG SONG
Bi 1. Cho tdin ABCD. Gi I, J, K, L theo th t l trungim ca cc cnh AB, BC ,CD ,DA Chng minh : IJ//KL vJK//IL .Bi 2. Cho tdin ABCD .Gi H, K l trng tm ca cc tamgic BCD v ACD .Chng minh rng HK//AB.Bi 3.Cho hnh chp S.ABCD c y l mt tgic li. Gi M,N ,E ,F ln lt l trung im ca cc cnh bn SC, SB, SC vSD.
1. Chng minh rng ME//AC , N F//BD2. Chng minh rng ba ng thng ME ,N F ,v SO(O l
giao im ca AC v BD) ng qui
3.
Chng minh rng 4 im M,N
,E,F ng phngBi 4.Cho hnh chp S.ABCD c ABCD l hnh bnh hnh. GiH, K l trung im SA, SB.
1. Chng minh rng HK//CD2.
Trn cnh SC ly im M. Dng thit din ca hnh chpvi mt phng (MKH).
Bi 5.Cho hnh chp S.ABCD y l hnh thang vi cc cnhy l AB v CD. Gi I, J lm lt l trung im ca DA v BC
v G l trng tm tam gic SAB.1. Tm giao tuyn ca (SAB) v (IJG)
Trng THPT N g Thi N him Bi tp ton 11
21
Bi 11. Chng minh rng ba sdng a, b, c lp thnh cp s
cng khi v chkhi cc s:baaccb +++
1,
1,
1lp
thnh cp scng.Bi 12.Tm bn shng lin tip ca mt cp scng bit tngca chng l 20 v tch ca chng l 348.
CP SNHN
Bi 1.Trong cc cp snhn di y, hy tnh shng nu
chra:1. 2; 1;
2
1;
4
1; ?7 =u
2. -3; 6; -12; 24; ?10 =u
3. 1;3
1;
9
1;
27
1; ?8 =u
Bi 2. Tm shng u, cng bi ca cc cp snhn, bit :
1.
=
=
192
96
6
5
u
u 2.
=+
=++
10
21
42
531
uu
uuu
3.
=
=+
240
90
62
53
uu
uu 4.
=
=
144
72
35
24
uu
uu
5.
=+
=+
325
65
71
531
uu
uuu 6.
=+
=+
20
10
653
542
uuu
uuu.
Bi 3.Tm cp snhn ( nu ) bit:1 2 3 4
2 2 2 21 2 3 4
15
85
u u u u
u u u u
+ + + =
+ + + =
Bi 4. Hy tm shng ca cp snhn, bit cp snhn :1.C 5 shng vi cng bi dng, shng thhai bng 3
v shng thtbng 6.
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Trng THPT N g Thi N him Bi tp ton 11
22
2. C 5 shng vi cng bi bng1
4shng thnht v
tng ca hai shng du bng 24.
Bi 5.Cho mt cp snhn c 7 shng, shng thtbng 6v shng thby gp 243 ln shng thhai. Hy tmcc shng cn li ca cp snhn .
Bi 6. Hy tm shng tng qut ca cp snhn ( nu ) c
=+
=+
123
16
43
52
uu
uu.
Bi 7.Tnh tng:
1. ...3
2.)1(...
9
4
3
21 1 +
+++= +
n
nS
2. ...1 32 +++= aaS vi21
1
+=a
Bi 8.Tnh tng tt ccc shng ca cp snhn (un) bit:1
2
2
2
64 2n
u
u
u
=
=
=
Bi 9.Mt cp scng v mt cp snhn u l cc dy tng.Cc shng thnht u bng 3, cc shng thhai bng
nhau. Tsgia cc shng thba ca cp snhn v cp scng l 9/5 .Tm hai cp sy.Bi 10. Tm hai sa, b bit rng 1,a,b l cp scng v 1,a2,b2l cp snhn.
Trng THPT N g Thi N him Bi tp ton 11
43
Bi 13: Cho hnh chp S.ABCD y l hnh thang, cnh y lnAB. Gi I, J, K ln lt l cc im nm trn SA, AB, CD
1.
Tm giao im ca IK v (SBD).
2.
Tm giao im ca SD v (IJK).3. Tm giao im ca SC v (IJK) .
THIT DIN
Bi 1: Cho tdin ABCD. Gi M, N ln lt l trung im cccnh AB v CD. P l im nm trn cnh AD nhng khng ltrung im. Tm thit din ca tdin ct bi mt phng(MN P).Bi 2: Cho tdin ABCD. Trn cc on AC, BC, BD ly ccim M, N , P sao cho MN khng song song vi AB, N P khngsong song vi CD. Xc nh thit din to bi mt phng(MN P) v tdin ABCD.Bi 6: Cho hnh chp SABCD. Gi M l 1 im thuc mintrong ca tam gic SCD.
1.
Tm giao tuyn ca hai mt phng (SBM) v (SAC).2. Tm giao
im c
a BM v m
t ph
ng (SAC).
3.
Tm thit din ca hnh chp ct bi mt phng (ABM).Bi 9: Cho hnh chp SABCD y l hnh bnh hnh tm O.Mt im M trn cnh SD sao cho SD = 3SM.
1. Tm giao tuyn ca (SAC) v (SBD).2. Xc nh giao im I ca BM v (SAC). Chng t I l
trung im ca SO.3. nh thit din ca hnh chp SABCD v (MAB).
Bi 14: Cho tdin ABCD ; im I nm trn BD v ngoiBD sao cho ID = 3IB; M; N l hai im thuc cnh AD; DC sao
cho MA=2
1 MD; N D =2
1N C
1. Tm giao tuyn PQ ca (IMN ) vi (ABC) ?2. Xc dnh thit din to bi (IMN ) vi tdin ?3.
Chng minh MN ; PQ ; AC ng qui ?Bi 17: Hnh chp SABCD c y ABCD l hnh thang vi AB
l y . Gi M ; N l trung im SB ; SC .1. Tm giao tuyn ca (SAD) v (SBC) ?
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Trng THPT N g Thi N him Bi tp ton 11
42
Bi 6: Cho hnh chp SABCD c ABCD l hnh thang vi yln l AB. Gi I, J ln lt l trung im ca SA, SB. M l imtu trn cnh SD.
1.
Tm giao tuyn ca(SAD) v (SBC).2. Tm giao im K ca IM vi mt phng (SBC).3.
Tm giao im N ca SC vi mt phng (IJM).Bi 7: Cho hnh chp SABCD c y l hnh bnh hnh. Gi Ml trung im ca SC.
1.
Tm giao im I ca ng thng AM vi mt phng(SBD).
2. Chng minh IA= 2IM.
3.
Tm giao im F ca SD v (ABM).4. im N thuc AB. Tm giao im ca MN v (SBD).
Bi 8: Cho tgic ABCD nm trong mt phng (P) c hai cnhAB v CD khng song song. Gi S l im nm ngoi (P) v Ml trung im ca on SC.
1. Tm giao im N ca SD v (MAB)2. Gi O l giao im ca AC v BD . CMR: SO, AM, BN
ng quiBi 9: Cho tdin ABCD. Hai im M, N ln lt nm trongtam gic ABC v tam gic ABD. I l im tu trn CD. Tmgiao ca (ABI) v ng thng MN .Bi 10: Cho hnh chp SABCD. Gi I, J l hai im trn cnhAD, SB
1. Tm cc giao im K, L ca IJ v DJ vi (SAC)2. AD ct BC ti O; OJ ct SC ti M. Chng minh A, K, L,
M thng hngBi 11: Cho tdin ABCD. Gi M, N ln lt l trung im caAC, BC. K l im trn cnh BD v khng trng vi trung imca BD.
1.
Tm giao im ca CD v (MN K).2. Tm giao im ca AD v (MN K)
Bi 12: Cho tdin ABCD. M, N l 2 im trn cnh AC, AD.O l 1 im bn trong BCD. Tm giao im ca:
1.
MN v (ABO).2. AO v (BMN ).
Trng THPT N g Thi N him Bi tp ton 11
23
CHNG IV. GII HN
GII HN CA DY S
Bi 1. Tnh cc gii hn sau:
1. lim92
142
2
+
n
nn 2. lim
37
76
652
+
+
nn
nnn
3. limnn
n
108
25 +
+ 4.
36
4325
4
+
+
nn
nn
5. lim 23
4
11100
3373
nn
nn
+
6. lim )32(3
)31(23
22
nn
nn
+
7. lim23
32
)42(
)2()23(
n
nn
8. lim
7
323
432
)5()51(
nn
nn
+
+
Bi 2.Tnh cc gii hn sau:
1.1
1lim
+
+
n
n 2.
2lim
3 3
+
+
n
nn
3.32
232lim
2
4
+
+
nn
nn 4.
12
21lim
2
+
+
n
nn
5. lim756
14
33 62
+
+++
nn
nnn 6.
12lim
43
+
++
n
nnn
7.
nnn
nn
+
++4 3
2 1lim 8.
23
11lim
2
+
++
n
nn
Bi 3.Tnh cc gii hn sau:
1.12
13lim
+n
n
2.n
nn
5.37
5.23lim
+
3.11
5)3(
5)3(lim
++
+
+nn
nn
4. lim52
12
24.5
43++
++
nn
nn
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Trng THPT N g Thi N him Bi tp ton 11
24
Bi 4.Tnh cc gii hn sau:
1. lim )nn +12 2. lim( 3n n+ )
3. )nnn ++ 1lim2
4. 12
1lim ++ nn
5. )nnn + 1lim 2 6. )nnn +3 32lim7. ( )3 31lim nn + 8.
++ nnnnlim
GII HN CA HM S
Bi 1.Tnh cc gii hn sau:
1.2
3lim
3
2
1 +
x
x
x
2.622
35lim
23
2
2 +++
++ xxx
xx
x
3.72
15lim
1 +
x
x
x
4. 32
4
2 2
232lim
+
++ xx
xx
x
Bi 2.Tnh cc gii hn sau:
1. 3 152lim2
3
+ x
xxx 2. 5
152lim
2
5 +
+ x
xx
x
3.
31 1
3
1
1lim
xxx 4.
253
103lim
2
2
2
+ xx
xx
x
5.xx
xx
x 4
43lim
2
2
4 +
+
6.6)5(
1lim
3
1 +
xx
x
x
7.6
44lim
2
23
2
++
xx
xxx
x
8.6
23lim
2
23
2
++
xx
xxx
x
9.6
293lim
3
23
2
+ xx
xxx
x
10.32
1lim
2
4
1 +
xx
x
x
Bi 3.Tnh cc gii hn sau:
1. .2
35lim
2
2
+ x
x
x 2.
2
153lim
2
x
x
x
3. 11lim0 + x
x
x 3. x
x
x
5
5lim5
Trng THPT N g Thi N him Bi tp ton 11
41
Bi 3.Cho hnh chp S.ABCD y ABCD l hnh bnh hnh ; Ol giao im hai ng cho; M ; N ln lt l trung im SA;
SD. Chng minh ba ng thng SO; BN ; CM ng quy.
GIAO IM CA NG THNG V MT PHNG
Bi 1: Cho tdin ABCD. Gi M, N ln lt l trung im caAC v BC. Gi K l mt im trn cnh BD khng phi l trungim. Tm giao im ca:
1.
CD v mt phng (MN K)2. AD v mt phng (MN K)
Bi 2: Cho tdin ABCD. Trn cc cnh AB v Ac ln lt lycc im M, N sao cho MN khng song song vi BC. Gi O lmt im nm trong tam gic BCD.
1. Tm giao im ca MN v (BCD)2.
Tm giao tuyn ca (OMN ) v (BCD)3. Mt phng (OMN ) ct cc ng thng BD v CD ti H
v K. Xc nh cc im H v K.Bi 3: Cho hnh chp SABCD. Gi I, J, K ln lt l cc imtrn cc cnh SA, AB, BC. Gi s ng thng JK ct ccng thng AD, CD ti M, N . Tm giao im ca cc ngthng SD v SC vi mt phng (IJK).Bi 4: Cho tdin ABCD. Gi M, N , P l cc im ln lt trncc cnh AC, BC, BD.
1.
Tm giao im ca CP v (MN
D).2.
Tm giao im ca AP v (MN D).Bi 5: Cho 4 im A, B, C, D khng ng phng. Gi M, N lnlt l trung im ca AC v BC. Trn BD ly im P sao choBP=2PD.
1. Tm giao im ca cc ng thng CD vi mtphng(MN P)
2. Tm giao im ca hai mt phng (MN P) v (ACD).
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Trng THPT N g Thi N him Bi tp ton 11
40
SE, SB ln lt ti M, N . Mt mt phng (Q) qua BC ct SD v
SA ln lt ti H v R.
1.
Gi I l giao im ca AM v DN , J l giao im ca
BH v ER. CMR bn im S, I, J, G thng hng.
2. Gi sK l giao im ca AN v DM, L l giao im
ca BR v EH. CMR ba im S, K, L thng hng.
Bi 9: Cho A; B; C khng thng hng ngoi mt phng ( ) .
Gi M; N ; P ln lt l giao im AB; BC; AC vi . Chngminh M; N ; P thng hng ?Bi 10: Hnh chp S.ABCD c y ABCD l hnh thang hai yl AD; BC. Gi M; N l trung im AB; CD v G l trng tmSAD. Tm giao tuyn ca :
1. (GMN ) v (SAB)2. (GMN ) v (SCD)3.
Gi giao im ca AB v CD l I; J l giao im ca haigiao tuyn cu a v cu b. Chng minh S; I; J thng hng .
4.
CHNG MINH BA NG THNG NG QUI
Bi 1.Cho hnh thang ABCD (AB// CD) im S nm ngoi mtphng cha ABCD. Gi M, N ln lt l trung im ca SC,SD. Gi I l giao im ca AD v BC, J l giao im ca AN vBM.
1.
CMR : S, I, J thng hng.2.
Gi O l giao im ca AC v BD. CMR : SO, AM, BN ng quy.Bi 2.Cho tdin ABCD. M, N ln lt l trung im BC, BD.
Cc im P v S ln lt thuc AD, AC sao cho1
3AR AD= ;
1
3
AS AC= . CMR : ba ng thng AB, MS, N R ng quy.
Trng THPT N g Thi N him Bi tp ton 11
25
5.25
34lim
25
+ x
x
x 6.
x
xx
x
11lim
2
0
++
7. ( )xxxx
x++ 121lim
2
0 8. xxx
x 3361lim 21 ++
+
9.1
132lim
21
+ x
xx
x
10.23
2423lim
2
2
1 +
xx
xxx
x
Bi 4.Tnh cc gii hn sau:
1.x
xx
x
+
55lim
0 2.
x
xxx
x
11lim
2
0
+++
3.x
xx
141lim3
0+
4.
x
xx 3
11lim3
0+
5.11
lim30 + x
x
x 6.
x
x
x
+ 51
53lim
4
7.1
lim2
1
x
xx
x 8.
23
1lim
2
3
1 +
+ x
x
x
9.x
xxx
3
0812lim
10.
157lim
23
1 +
xxx
x
Bi 5.Gii hn mt bn:
1.1
3 2lim
1x
x
x+
2.
34
1lim
2
4
3 ++
++ xx
x
x
3. )(lim1
xfx
bit ( )
>+
=
1;1
1;13
2 xx
xxxf
4. )(lim1
xfx
; )(lim3
xfx
bit ( )
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Trng THPT N g Thi N him Bi tp ton 11
26
1.x
x
x
5sinlim
0 2.
x
x
x 3
2tanlim
0
3.2
0
cos1lim
x
x
x
4.x
xx
x 30 sin
sintanlim
5.x
x
x 2sin
121lim
0
+
6.30 45
sin.3sin.5sinlim
x
xxx
x
7.30
tansinlim
x
xx
x
8.xx
x
x sin
cos1lim
3
0
9.x
xx
x sin
112lim
3 2
0
++
10.2
2
0
cos1lim
x
xx
x
+
Bi 7.Tnh cc gii hn sau:
1.32
3
662
13lim
xx
xx
x
++
2.3
2x
x 5lim
x 1+
+
3.2
5 3
2lim
3 4 1x
x
x x+
+ 4.
2
5 1lim
3 4 1x
x
x x
+ +
5.2
3
(2 1) (1 3 )lim
( 1)x
x x
x
+ 6. ( ) ( )
( )50
3020
12
2332lim
+
+
x
xx
x
Bi 8. Tnh cc gii hn sau:
1. xxx
+
1lim 2 2. 2lim ( 3 )x
x x x
+
3. )xxxxx
914lim 22 ++
4. 32 2lim ( 3 )x
x x x
+
5. 34412lim 2 ++++
xxxx
6. )13lim 3 23 ++
xxxxx
7. 2lim ( 2)x
x x x
+ + + 8. xxxxx
22lim 23 23 ++
Bi 9.Tnh cc gii hn sau:
1.0
lim . cotx
x x
2.1
lim(1 ).tan2x
xx
3.3
lim(4 ).sinx
xx
+ 4.0
lims in2 .cot6x
x x
Trng THPT N g Thi N him Bi tp ton 11
39
Bi 7: Cho tdin SABC. Gi M,N l cc im trn cc onSB v SC sao cho MN khng song song vi BC . Tm giaotuyn ca mt phng (AMN ) v (ABC), mt phng (ABN ) v
(ACM).Bi 8: Cho tdin ABCD. Trn AB ly M vi AM =
3
1AB. Gi
I, K ln lt l trung im ca AC, AD. nh giao tuyn (d) camt phng (MIK) v (BCD).Bi 9: Cho tdin SABC. Gi I, J, K l ba im tu trn SB,AB, BC sao cho JK khng song song vi AC v SA khng songsong vi IJ. nh giao tuyn ca (IJK) v (SAC).Bi 10: Cho t din ABCD. Gi E, F, G l ba im trn AB,AC, BD sao cho (EF) ct (BC) ti I , (EG) ct (AD) ti H. nhgiao tuyn ca mt phng (EFG) vi hai mt phng (BCD) v(ACD).
CHNG MINH BA IM THNG HNG.
Bi 1: Cho tdin SABC. Trn SA, SB v SC ln lt ly ccim D, E, F sao cho DE ct AB ti I, EF ct BC ti J, FD ct
CA ti K. Chng minh rng ba im I, J, K thng hng.
Bi 2: Cho hai mt phng (P) v (Q) ct nhau theo giao tuyn d.
Trong (P) ly hai im A v B sao cho AB ct d ti I. O l mt
im nm ngoi (P) v (Q) sao cho OA v OB ln lt ct (Q)
ti A
v B
.1. Chng minh ba im I, A, Bthng hng.
2. Trong (P) ly im C sao cho A, B, C khng thng hng.
GisOC ct (Q) ti C, BC ct BCti J, CA ct CAti K.
Chng minh I, J, K thng hng.
Bi 8: Cho tdin SABC c D, E ln lt l trung im AC,
BC v G l trng tm tam gic ABC. Mt phng (P) qua AC ct
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38
CHNG II. QUAN HSONG SONG
TM GIAO TUYN CA HAI MT PHNG.
Bi 1: Cho S l mt im khng thuc mt phng hnh bnhhnh ABCD. Tm giao tuyn ca hai mt phng (SAC) v(SBD).Bi 2: Trong mt phng ( ) cho t gic ABCD sao cho AB,
CD khng song song. S l im nm ngoi ( ) . Tm giao tuynca cc cp mt phng sau:
1.
(SAC) v (SBD).2. (SAB) v (SCD).
Bi 3: Trong mt phng ( ) cho ba im A, B, C. S l im
khng thuc ( ) . M, N , I ln lt l trung im ca AB, BC,SA.
1. Tm giao tuyn ca (SAN ) v (SCM).
2.
Tm giao tuyn ca (SCM) v (BIC).Bi 4: Cho t din ABCD v im M thuc min trong caACD . Gi I v J tng ng l hai im trn cnh BC v BD
sao cho IJ khng song song vi CD.1. Hy xc nh giao tuyn ca hai mt phng (IJM) v
(ACD).2. Ly N l im thuc min trong ca ABD sao cho JN
ct on AB ti L. Tm giao tuyn ca hai mt phng (MN J) v
(ABC).Bi 5: Cho hnh chp S.ABCD c y l tgic ABCD c haicnh i din khng song song. Ly im M thuc min trongca SCD . Tm giao tuyn ca hai mt phng:
1. (SBM) v (SCD).2. (ABM) v (SCD).3. (ABM) v (SAC).
Bi 6: Cho tdin ABCD. M v N ln lt l trung im AD v
BC. Tm giao tuyn ca hai mt phng (MBC) v (N AD).
Trng THPT N g Thi N him Bi tp ton 11
27
HM SLIN TC
Bi 1. Xt tnh lin tc cc hm ssau ti 0x :
1. ( )3 11
2 1
+ =
=
xkhi x
f x x
khi x
ti 0 1x =
2. ( )
3 82
25 2
= =
xkhi x
f x x
khi x
ti 0 2x =
3.2
55
( ) 2 1 3
( 5) 5
= =
xkhi x
f x x
x khi x
ti x0= 5
4.
3
2
1 cos0
sin( )
1 02
xkhi x
xf x
khi x
=
=
ti x0= 0
5. ( )
2 22
25 2
>
=
x xkhi x
f x x
x khi x
ti 0 2x =
6. ( )
11
2 12 1
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28
8.
2
2
3 21
1
1( ) 14
11
6 7
+ >
= =
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36
Bi 2.Cho hnh chnht ABCD. Gi E,F,H,I ln lt l trungim ca AB,CD,BC,EF. Hy tm mt php di hnh bin tamgic AEI thnh tam gic FCH.
PHP VNT
Bi 1.Xc nh nh ca im A(4,-5) qua php vt tm I(-2;6), ts-2.Bi 2. Cho im M(-1;5) v ng thng d: 2x-3y-8=0. Xcnh nh ca M v d qua php vttm O tsbng 2.
Bi 3. Cho im I(2;-1) v im J(7:4). Tm tm v t ca 2ng trn (C)(I;2) v ng trn (C)(J;3).Bi 4.Cho tam gic OMN . Dng nh ca M, N qua php vttm O, tsk trong cc trng hp sau:
1. 3k= 2.1
2k= 3.
3
4k=
Bi 5.Tm php vt bin:
1.
: 12 4x yd = thnh ' : 2 6 0d x y = .
2. 2 2( ) : ( 4) 2C x y+ + = thnh 2 2( ') : ( 2) ( 3) 8C x y + =
PHP N G DN G
Bi 1.Cho im A(3;-4) v ng thng d: 9x+y-6=0 . Vit pt
ng thng d l nh ca d qua php ng dng c c bngcch thc hin lin tip php Ov php V(A,1/3).Bi 2.Cho ng trn (C) c tm I(-1;3), bn knh bng 2. Vitphng trnh ng trn nh ca (C) qua php ng dng cc tvic thc hin lin tip php V(O,3)v php Oy.Bi 3.Cho hnh vung ABCD tm O, M l trung im cnh AB.Xc nh php ng dng bin OAM thnh DBC.
BI TP N CHNG
Trng THPT N g Thi N him Bi tp ton 11
29
3
12
4( )
3 2 2 22
ax khi x
f x
x khi xx
+
=+
>
Bi 5. Chng minh phng trnh:1. 22 6 1 0 + =x x c t nht 2 nghim.2. 32 10 7 0 =x x c t nht hai nghim.3. cos 0 =x x c nghim.
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CHNG IV. O HM
Bi 1.Tnh o hm cc hm ssau:1. 5 3 24= +y x x x x 2.
36= +y x
x
3. 21 1
= + +y xx x
4.3
2 3= + +y x
x x
5. ( )( )3 2 24 2 7= y x x x x 6. 2( 2) 1= +y x x
7.2 1
=+
xy
x 8. 2 3
4=
+
xy
x
9.25 3
2
=
x xy
x 10.
1
1
=
+
xy
x
11. ( )37 25= y x x 12. ( )
52 33 4= + +y x x x
Bi 2. Tnh o hm cc hm ssau:
1.sin
= x
yx
2. cos1
=+
xy
x
3.sin cos
sin cos
+=
x xy
x x 4. sin3 cos tan
5= + +
xy x x
5.
( )
102 sin5= y x x 6. 3sin 3=y x
7. 2cot 1= +y x x 8. ( )2sin cos 2=y x
9. 2 sin 3=y x x 10. 3 2sin 1= +y x
11. 2 2tan 3 cot 2= +y x x 12. sin 4 cos .= +y x x x Bi 3.Gii cc bt phng trnh:
1. ( )' 0>
f x vi ( )
2 5 4
2
x x
f x x
+=
Trng THPT N g Thi N him Bi tp ton 11
35
PHP QUAY
Bi 1.Cho hnh vung ABCD c tm O. Tm nh ca tam gic
OAB qua php quay tm C gc quay -900
Bi 2. Tm to cc im nh ca A(-3;4), B(-5;1), C(-2;3)qua php quay Q(O,90
o)
Bi 3. Cho im M(3;-4) v ng thng d: 6x-y+10=0. Xcnh nh ca M v d qua php quay tm O mt gc 900.Bi 4. Trong mt phng Oxy, tm php quay Q bin im A(-1,5) thnh im B(5,1).Bi 5. Trong mt phng Oxy, cho im A(0,3). Tm
B = Q (A)(O ; 45 )
Bi 6. Trong mt phng Oxy, cho ng trn2 2(C) : (x 3) (y 2) 4 + = . Tm (C ) = Q (C)
(O ; 90 )
Bi 7.Trong mt phng ta Oxycho ng trn c phng
trnh : 2 2 2 4 4 0x y x y+ + = . Vit phng trnh ng trn
l nh ca ng trn cho qua php quay tm O gc quay090 ; - 090
PHP DI HN H
Bi 1. Cho 2 im A(-2;1), B(3;5) v ng thng d: 4x-9y+6=0.
1. Xc nh nh ca im A qua php di hnh c c bngcch thc hin lin tip php Q(O,90o)v php B.
2. Xc nh nh ca im B qua php di hnh c c bngcch thc hin lin tip php Av Oy.
3. Xc nh nh ca im A qua php di hnh c c bngcch thc hin lin tip php Q(O,90
o)v php d.
4. Xc nh nh ca d qua php di hnh c c bng cchthc hin lin tip php Ov tnh tin
ABT
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Trng THPT N g Thi N him Bi tp ton 11
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Bi 1. Trong mt phng Oxy, hy tm nh ca im M( 2, 1)qua php i xng trc Ox, ri i xng trc Oy.
Bi 2.Trong mt phng Oxy, cho im A(-5; 6), ng thng d:2x-3y-1=0 v ng trn (C): (x-1)2+(y+2)2= 25.1. Xc nh nh ca A v ng thng d qua php Ox.2. Xc nh ng trn (Co) sao cho (C) l nh ca (Co)
qua php Oy.3. Xc nh nh ca (C) qua php d.
Bi 3. Cho im M(2;-7) v ng cong (C) c phng trnhy = x3+3x2-2x+1 .
1.
Xc nh tocc nh ca im M qua php Ox, Oy.2. Vit phng trnh ng cong (C) l nh ca (C) qua
php Ox.Bi 4. Trong mt phng Oxy, cho ng thng( ) : x 5y 7 = 0 + v ( ) : 5x y 13 = 0 . Tm php ixng qua trc bin ( ) thnh ( ) .Bi 5. Cho tgic ABCD. Goi O l giao im ca AC v BD.
Xc nh nh ca tam gic AOB qua php i xng trc CD.PHP I XN G TM
Bi 1.Cho 2 im M(-2;9), N (1;4). Xc nh cc im M1,M2ln lt l nh ca M qua php O , N .Bi 2. Cho im I(-4;3), ng thng d: x-2y+5=0 v ngtrn (C):x2 + y2 -2x+6y+1=0.
1.
Xc nh nh ca I, d v (C) qua php O.2. Vit phng trnh ng thng D l nh ca D quaphp I.
3. Vit phng trnh ng trn (C) l nh ca (C) quaphp I.
Bi 3. Chng minh rng2 2
2 2( ) : 1
x yE
a b+ = v
2 2
2 2( ) : 1
x yH
a b =
c tm i xng l gc ta O.
Trng THPT N g Thi N him Bi tp ton 11
31
2. ( )' 0g x vi ( ) 22 1
1
xg x
x
=
+
3. ' 0y vi
2
2
3
1
xy x
+= +
4. ' 0y vi2
2 1
4
xy
x x
=
+ +
Bi 4.Chng minh rng:1. Hm s xy tan= tha hthc: .01 '2 =+ yy
2. Hm s xy 2cot= tha hthc : .022 '2 =++ yy
3. Hm s 34
xyx
=+
tha hthc : ( ) ( )2
2 ' 1 ''y y y= .
Bi 5.Tnh o hm cp hai ca cc hm ssau:1. 2 cosy x x x= +
2. ( )2 1 tany x x= +
3. 2cos=y x
4. 4 cos2= y x x Bi 6. Vit phng trnh tip tuyn ca thhm s:
1.1
2
=
+
xy
xti im c honh bng 4.
2.2 2 6
1
+ =
x xy
xbit c honh tip im l 3.
3. 3 21 13= + +y x x x bit hsgc k = -3.
4.2 2
2
=
+
x xy
xbit tip tuyn song song vi ng thng
xy 32 = .
5. 3 21
13
= + +y x x x bit tip tuyn vung gc vi ng
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thng: 54
1+= xy .
6. 4 23 4y x x= bit tip tuyn i qua im A(0, -4).
Bi 7. Cho hm s113
+=
x
xy c thl (C). Vit phng trnh
tip tuyn ca (C) sao cho tip tuyn :
1. C tung tip im l 2.2. Vung gc vi ng thng: 4 10y x= + .
Bi 8.Qua im4 4
( , )9 3
A c thkc bao nhiu tip tuyn
n thhm s3
22 33
xy x x= + . Vit phng trnh cc tip
tuyn .
Trng THPT N g Thi N him Bi tp ton 11
33
PHN II. HNH HC
CHNG I. PHP DI HNH V PHP NG DNGTRONG MT PHNG
PHP TNN H TIN
Bi 1.Trong mt phng Oxy, cho im A(-1,2), B(0,1), C(3,-1)v vect ( 2,3)v=
. Hy tm nh ca cc im trn qua php
tnh tin theo vect v .Bi 2.Trong mt phng Oxy, cho ng thng d: 2x 3y +1 =0 v ng trn (C): x2 + y2- 2x - 4y 4 =0. Hy tm nh ca dv (C) qua php tnh tin theo vect (1, 2)v=
.
Bi 3.Trong mt phng Oxy, cho ng thng ct Ox ti A(-1, 0) v ct Oy ti B(0 ,2). Hy tm nh ca qua php tnhtin theo vect u = (2; 1)
.
Bi 4.Hy tm nh ca ng trn 2 2(C) : (x 3) (y 2) 1 + + = qua php tnh tin theo vect u = ( 2;4)
.
Bi 5. Trong mt phng Oxy, cho A( 5;2) , C( 1;0) . Bit
B = T (A) , C = T (B)u v . Tm u va v
c thbin A thnh C.
Bi 6.Cho ABC c trng tm G. Dng nh ca :1. on thng AB qua php tnh tin
GCT
2. ABC qua php tnh tin 2AGT
Bi 7.Cho hnh bnh hnh ABCD c tm O. Xc nh :1. nh ca ABD qua php tnh tin
3OCT
2. im E sao cho php tnh tinAC
T bin E thnh D.
PHP I XN G TRC
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