Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu

CNG N TP HC K II, MN TON LP 11

NM HC 2010 - 2011
A. I S & GII TCH
I. CC DNG BI TP THNG GP
CHNG IV : GII HN
1/ Chng minh dy s (u
n
) c gii hn 0.
Phng php: - Vn dng nh l: Nu |u
n
| v
n
, n v lim v
n
= 0 th limu
n
= 0
- S dng mt s dy s c gii hn 0:
lim 0
1
n

,
lim 0
1
n

,
3
lim 0
1
n

,
lim 0
n
q
vi |q| < 1
2/ Tm gii hn ca dy s, ca hm s.
Phng php: Vn dng cc nh l v gii hn hu hn v cc quy tc tm gii hn v cc
- Cc quy tc tm gii hn v cc ca dy s:
+) Nu limu
n
= + th
lim 0
1
n
u

- Cc quy tc tm gii hn v cc ca hm s:
+) Nu
( )
0
lim
x x
f x

+
th
( )
0
lim 0
1
x x
f x

- Ch khi gp cc dng v nh:

0
; ; ; 0.
0

ta phi kh cc dng v nh bng cch: chia t v

mu cho n hoc x m ln nht; phn tch t hoc mu thnh nhn t n gin, nhn c t v mu
vi mt lng lin hp;
3/ Tnh tng ca cp s nhn li v hn
limu
n
limv
n
= L lim(u
n
v
n)
+ L >0 +
+ L < 0

L >0

L < 0 +
limun=L limvn
Du ca
v
n
lim
n
n
u
v
L >0
0
+ +
L > 0 -

L < 0 +

L < 0 - +
) ( lim
0
x f
x x
) ( lim
0
x g
x x
) ( ). ( lim
0
x g x f
x x
+ L > 0 +
- -
+ L < 0 -
- +
) ( lim
0
x f
x x
) ( lim
0
x g
x x
Du ca
g(x) ) (
) (
lim
0 x g
x f
x x
L > 0
0
+ +
- -
L < 0
+ -
- +
1
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
Cho CSN (u
n
) li v hn (vi
1 < q
), ta c :

1
1 1 1
1
n
u
S u uq uq
q
+ + + +

L L
4/ Xt tnh lin tc ca hm s
Phng php: Xt tnh lin tc ca hs f(x) ti x
0
:
+) Tnh f(x
0
)
+) Tm
( )
0
lim
x x
f x

(nu c)
- Nu
( )
0
lim
x x
f x

khng tn tif(x) gin on ti x

0
.
- Nu
( ) ( )
0
0
lim
x x
f x L f x

f(x) gin on ti x
0
- Nu
( ) ( )
0
0
lim
x x
f x L f x

f(x) lin tc ti x
0.
5/ Chng minh s tn ti nghim ca mt phng trnh.
Phng php: Vn dng h qu ca nh l v gi tr trung gian: Nu hm s y = f(x) lin tc trn on [a;b]
v f(a).f(b) < 0 th phng trnh f(x) = 0 c t nht 1 nghim nm trong (a ; b).
CHNG V: O HM
1/ Tm o hm ca hm s
Phng php: p dng cc cng thc tnh o hm
+) Cc quy tc tnh o hm:
'
2
'
2
( )' ' '
( . )' '. '.
( . )' . '
'. '.
1 '
u v u v
u v u v v u
k u k u
u u v v u
v v
v
v v
t t
+

,
_

,

( )
( )
( )
1
'
2
' 0 ; ' 1
' .
1 1
1
'
2
n n
c x
x n x
x x
x
x

_


,

( )
( )
1
'
2
' . . '
1 '
'
'
2
n n
u n u u
u
u u
u
u
u

_


,

+) o hm ca hm hp: Nu [ ( )] y f u x th
' ' '
.
x u x
y f u
+) o hm ca cc hm s lng gic:
( )
( )
( )
2
2
sin ' cos
cos ' sin
1
tan '
cos
1
(cot ) '
sin
x x
x x
x
x
x
x

( )
( )
( )
2
2
sin ' '.cos
cos ' '.sin
'
tan '
cos
'
(cot ) '
sin
u u u
u u u
u
u
u
u
u
u

2/ Vit phng trnh tip tuyn ca th hm s.
Phng php:pt tip tuyn ca th hm s y = f(x) ti im M
0
c honh x
0
c dng:
y = f(x
0
) (x x
0
) + f(x
0
)
3/ Vi phn
2
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
- Vi phn ca hm s ti nt im:
0 0
( ) '( ). df x f x x
- ng dng vi phn vo tnh gn ng:
0 0 0
( ) ( ) '( ) f x x f x f x x + +
- Vi phn ca hm s:
( ) '( ) df x f x dx
hay
' dy y dx
4/ o hm cp cao:
- o hm cp hai ca hm s: f= (f).
- o hm cp n ca hm s: f
(n)
= [f
(n-1)
].
II. BI TP
CHNG IV: GII HN
Bi 1: Chng minh cc dy s sau c gii hn 0:
( )
2
1
)
2 1
n
n
a u
n

+
sin 2
)
1
n
n
b u
n

+
2
cos3
)
n
n n
c u
n n
+

+
cos
)
1
n
n
d u
n n

+
( )
1
1
)
3
n
n n
e u
+

2
)
3 1
n
n n
f u
+
( )
1 1
1
1
)
3 5
n
n n n
g u
+ +

+
) 1
n
h u n n +
Bi 2: Tm cc gii hn sau:
3
3 2
2 3 1
) lim
n n
a
n n
+
+
3
2
3 2
) lim
2 1
n n
b
n
+
+

3
3 2
) lim
2 1
n
c
n n
+
+
5
3 2
1 2 3
) lim
( 2) (5 1)
n n
d
n n
+


2
4 1
) lim
1 2
n n
e
n
+ +

3 2.5
) lim
3.5 4
n n
n n
f

3 4 1
) lim
2.4 2
n n
n n
g
+
+
2 2
4 1 9 2
) lim
2
n n
h
n
+ +

) lim
n
i u
vi
( )
1 1 1 1
...
1.2 2.3 3.4 1
n
u
n n
+ + + +
+
S: a) -3 b) + c) 0 d) -3/25 e) -1 f) -2/3 g) -1/2 h) 1 i) 1
Bi 3 : Tnh cc gii hn sau:
2
) lim(3 1) a n n +
4 2
) lim( 2 3) b n n n + + ( )
2
) lim 3 sin 2 c n n n +
2
) lim 3 1 d n n +
( )
) lim 2.3 5.4
n n
e
2
) lim 3 1 2 f n n +
2
) lim 1 g n n + ( )
+
2
)lim h n n n
( )
2
) lim 3 6 1 7 i n n n +
( )
) lim 1 k n n n
( )
2
) lim 3 l n n n
( )
3 3 2
) lim m n n n +
S: a) + b) - c) + d) + e) - f) - g) 0 h) + i) - k) -1/2 l) -3/2 m) 1/3
Bi 4: Tnh tng ca cp s nhn li v hn sau:
a)
1
1 1 1 1
1, , , ,..., ,...
2 4 8 2
n
_


,
b)
1
1 1 1 1
1, , , ,..., ,...
3 9 27 3
n
_

,
S: a) 2/3 b) 3/2
Bi 5: Tm gii hn ca cc hm s sau: (Dng

):
a)
3
3 2
5 1
lim
2 3 1
x
x x
x x
+
+
+ +
b)
3
3 2
lim
2 1
x
x
x

+
+
c)
3 2
2
5 1
lim
3
x
x x
x x

+
+

d)
5 3
2 3
2 4
lim
1 3 2
x
x x x
x x
+
+

2
3 2
5 1
) lim
2 3 1
x
x
e
x x
+

+ +
f)
2 2
2 4 1
lim
2 5
x
x x x
x

+ +

S: a) -1/2 b) - c) - d) - e) 0 f) -1/5
3
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
Bi 6: Tm gii hn ca cc hm s sau: (Dng: a.):
a)
3 2
lim( 2 3 1)
x
x x x

+ +
b)
4 3
lim( 5 3)
x
x x x
+
+ +
c)
2
lim 4 2
x
x x
+
+ +
d)
2
lim 3 2
x
x x

+
e)
( )
2
lim 3 2
x
x x x
+
+
f)
( )
2
lim 2
x
x x x

+ +
S: a) + b) - c) + d) + e) - f) +
Bi 7: Tm gii hn ca cc hm s sau: (Gii hn mt bn):
a)
3
1
lim
3
x
x
x

b)
( )
2
4
1
lim
4
x
x
x

c)
3
2 1
lim
3
x
x
x
+

d)
2
2 1
lim
2
x
x
x
+

+
+
e)
2
0
2
lim
x
x x
x x

f)
1
3 1
lim
1
x
x
x

+
S: a) - b) - c) +

d) +

e) 1 f) +

Bi 8: Tm gii hn ca cc hm s sau: (Dng

0
0
):
a/
2
3
9
lim
3
x
x
x

b/
2
1
3 2
lim
1
x
x x
x

c)
2
3
3
lim
2 3
x
x
x x

+
+
d)
3
2
1
1
lim
1
x
x
x

e)
2
2
1
2 3
lim
2 1
x
x x
x x

+

f)
2
2
lim
7 3
x
x
x

+
g)
2
3
9
lim
1 2
x
x
x

+
h)
4
2 1 3
lim
2
x
x
x

i)
1
2 1
lim
5 2
x
x
x

+
+
k)
2
2
3 2
lim
2
x
x x
x

S: a) 6 b) -1 c) -4 d) 3/2 e) 4/3 f) -6 g) 24 h) 4/3 i) 2 k) 0

Bi 9: Tm gii hn ca cc hm s sau: (Dng 0. ):
a)
0
1 1
lim 1
1
x
x x

+
,
b)
( )
2
1
2 3
lim 1
1
x
x
x
x
+

c)
2
3
2 1
lim 9.
3
x
x
x
x
+

d/
( )
3
2
2
lim 8
2
x
x
x
x

S: a) -1 b) 0 c) + d) 0
Bi 10: Tm gii hn ca cc hm s sau: (Dng - ):
a)
( )
2
lim 1
x
x x
+
+
b)
( )
2 2
lim 2 1
x
x x x
+
+ +
c)
( )
2
lim 4 2
x
x x x

+
d)
( )
2 2
lim 1
x
x x x

S: a) 0 b) 1 c) 1/4 d) 1/2
Bi 11: Tm gii hn ca cc hm s sau: (p dng
0
sin
lim 1
x
x
x

)
a)
0
sin3
lim
x
x
x

b)
2
0
sin sin 2
lim
3
x
x x
x

c)
2
0
1 cos
lim
sin
x
x
x x

d)
0
sin .sin 2 ....sin
lim
n
x
x x nx
x

S: a) 3 b) 2/3 c) 1 d) n!
Bi 12: Xt tnh lin tc ca cc hm s sau:
a)
2
4
-2
( )
2
4 -2
x
khi x
f x
x
khi x

+ '

ti x
0
= -2 b)
2
4 3
khi x<3
( )
3
5 khi 3
x x
f x
x
x
+

'

ti x
0
= 3
c)
2
2 3 5
1
( )
1
7 1
x x
khi x
f x
x
khi x
+
>

'

ti x
0
= 1 d)
2 1
3
( )
3
3 3
x
khi x
f x
x
khi x

'

ti x
0
= 3
e/
2
2
2
( ) 2
2 2 2
x
khi x
f x x
khi x

'

ti x
0
=
2
f)
2
2
( ) 1 1
3 4 2
x
khi x
f x x
x khi x

>

'

ti x
0
= 2
S: a) lin tc ; b) khng lin tc ; c) lin tc ; d) khng lin tc ; e) lin tc ; f) lin tc
Bi 13: Xt tnh lin tc ca cc hm s sau trn TX ca chng:
4
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
a)
2
3 2
2
( )
2
1 2
x x
khi x
f x
x
khi x
+

'

b) ( )
2
1
2
2 ( )
3 2
x
khi x
x f x
khi x

'

c) ( )
2
2
x 2
2
5 x 2
x x
khi
f x
x
x khi

>

'

d) ( )
2
2
0
0 1
2 1 1
x khi x
f x x khi x
x x khi x
<

<
'

S: a) hslin tc trn R ; b) hs lin tc trn mi khong (-; 2), (2; +) v b gin an ti x = 2.
c) hslin tc trn R ; d) hs lin tc trn mi khong (-; 1), (1; +) v b gin an ti x = 1.
Bi 14: Tm iu kin ca s thc a sao cho cc hm s sau lin tc ti x
0
.
a) ( )
2
2
1
1
1
x x
khi x
f x
x
a khi x

+ '

vi x
0
= -1 b)
2
1
( )
2 3 1
x khi x
f x
ax khi x
<

'

vi x
0
= 1
c)
7 3
2
( )
2
1 2
x
khi x
f x
x
a khi x

'

vi x
0
= 2 d)
2
3 1 1
( )
2 1 1
x khi x
f x
a khi x
<

'
+

vi x
0
= 1
S: a) a = -3 b) a = 2 c) a = 7/6 d) a = 1/2
Bi 15: Chng minh rng phng trnh:
a)
4
5 2 0 x x + c t nht mt nghim.
b)
5
3 7 0 x x c t nht mt nghim.
c)
3 2
2 3 5 0 x x + c t nht mt nghim
d)
3
2 10 7 0 x x c t nht 2 nghim.
e) cosx = x c t nht mt nghim thuc khong (0; /3)
f) cos2x = 2sinx 2 = 0 c t nht 2 nghim.
g)
3 2
3 1 0 x x + c 3 nghim phn bit.
h) ( ) ( )
3
2 2
1 1 3 0 m x x x + + lun c t nht 1 nghim thuc khong (-1; -2) vi mi m.
i) ( ) ( )
3
2 4
1 4 3 0 m x x x + lun c t nht 2 nghim vi mi m.
CHNG V : O HM
Bi 1: Dng nh ngha tm o hm cc hm s sau:
a)
3
y x b)
2
3 1 y x + c) 1 y x + d)
1
1
y
x

Bi 2: Tnh o hm cc hm s sau:
1) +
3 2
5
3 2
x x
y x 2) 3
2
2
5
+
x
x y 3) +
2 3 4
2 4 5 6
7
y
x x x x
4) ) 1 3 ( 5
2
x x y 5) y = (x
3
3x )(x
4
+ x
2
1) 6)
3 2
) 5 ( + x y 7)
) 3 5 )( 1 (
2 2
x x y + 8)
) 2 3 )( 1 2 ( + x x x y
9)
3 2
) 3 ( ) 2 )( 1 ( + + + x x x y
10) ( )
_
+

,
2
3 1 y x x
x
11)
3
2 y x
12) y = ( 5x
3
+ x
2
4 )
5

13)
4 2
3 y x x +
14) ( ) ( ) ( )
2
2 1 2 3 7 y x x x + +
15)
2
2 5
2
x
y
x

+
5
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
16)
2
1
2 3 5
y
x x

+
17)
3
2
2
1
x x
y
x x

+ +
18)
+ +

2
2
7 5
3
x x
y
x x

19)
7 6
2
+ + x x y
20) 2 1 + + x x y 21)
1 ) 1 (
2
+ + + x x x y
22)
1 2
3 2
2
+
+

x
x x
y
23)
1 x
y
1 x
+

24)
( )
3
2
2 3 1 y x x +
25)
( )
3
2 3
2 y x x x x + + 26) y = x (x
2
- x +1) 27)
3
2
2 3
2
x
y x x
x
_
+

,

Bi 3: Tnh o hm ca cc hm s sau:
1) y = 5sinx 3cosx 2) y = cos (x
3
) 3) y = x.cotx 4)
2
) cot 1 ( x y + 5)
x x y
2
sin . cos 6)
3
1
cos cos
3
y x x 7)
2
sin
4
x
y 8)
x x
x x
y
cos sin
cos sin

+
9)
3
y cot (2x )
4

+ 10)
2
sin (cos3 ) y x 11)
3 2
y cot 1 x +
12) x x y 3 sin . sin 3
2

13)
2
y 2 tan x +
14)
3
cosx 4
y cotx
3sin x 3
+ 15)
sin(2sin ) y x
16)
4
sin 3 y x p = -
17)
2 2
) 2 sin 1 (
1
x
y
+

18)
xsinx
y
1 tanx

+
19)
sinx x
y
x sinx
+
20) y 1 2tanx +
Bi 4: Cho hai hm s :
4 4
( ) sin cos f x x x + v
1
( ) cos 4
4
g x x
Chng minh rng:
'( ) '( ) ( ) f x g x x
.
Bi 5: Cho 2 3
2 3
+ x x y . Tm x : a) y > 0 b) y < 3
S: a)
0
2
x
x
<

>

b)
1 2 1 2 x < < +
Bi 6: Gii phng trnh : f(x) = 0 bit rng:
a) f(x) = cos x + sin x + x. b) f(x) = x x cos x sin 3 +
c) f(x) = 3cosx + 4sinx + 5x d) f(x) = 2x
4
2x
3
1
Bi 7: Cho hm s f(x) 1 x. Tnh: f(3) (x 3)f '(3) + +
Bi 8: a) Cho hm s:
2
2 2
2
+ +

x x
y . Chng minh rng: 2y.y 1 =y
2
b) Cho hm s y =
4 x
3 x
+

. Chng minh rng: 2(y)

2
=(y -1)y
c) Cho hm s
2
y 2x x . Chng minh rng: +
3
y y" 1 0
Bi 9: Chng minh rng
'( ) 0 f x x >
, bit:
a/
9 6 3 2
2
( ) 2 3 6 1
3
f x x x x x x + + b/
( ) 2 sin f x x x +
Bi 10: Cho hm s
2
2
x x
y
x
+

(C)
a) Tnh o hm ca hm s ti x = 1.
b/ Vit phng trnh tip tuyn ca (C) ti im M c honh x
0
= -1.
B i 11: Cho hm s y = f(x) = x
3
2x
2
(C)
a) Tm f(x). Gii bt phng trnh f(x) > 0.
6
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
b) Vit phng trnh tip tuyn ca (C) ti im M c honh x
0
= 2.
c) Vit phng trnh tip tuyn ca (C) bit tip tuyn song song vi ng thng d: y = - x + 2.
Bi 12: Gi ( C) l th hm s :
3 2
5 2 y x x + . Vit phng trnh tip tuyn ca (C )
a) Ti M (0;2).
b) Bit tip tuyn song song vi ng thng y = -3x + 1.
c) Bit tip tuyn vung gc vi ng thng y =
1
7
x 4.
Bi 13: Cho ng cong (C):
2
2
x
y
x
+

. Vit phng trnh tip tuyn ca th (C)

a) Ti im c honh bng 1
b) Ti im c tung bng
1
3
c) Bit tip tuyn c h s gc l 4
Bi 14: Tnh vi phn cc hm s sau:
a) 1 2
3
+ x x y b)
2
sin
4
x
y c)
7 6
2
+ + x x y
d) x x y
2
sin . cos e)
2
) cot 1 ( x y +
Bi 15: Tm o hm cp hai ca cc hm s sau:
1)
1
2
x
y
x
+

2)
2
2 1
2
x
y
x x
+

+
3)
2
1
x
y
x

4)
2
1 y x x +

5)
2
sin y x x 6)
2
(1 ) cos y x x 7) y = x.cos2x 8) y = sin5x.cos2x
S: 1)
( )
3
6
''
2
y
x

2)
( )
3 2
3
2
4 10 30 14
''
2
x x x
y
x x
+ +

+
3)
( )
( )
2
3
2
2 3
''
1
x x
y
x
+

4)
( )
3
2 2
2 3
''
1 1
x x
y
x x
+

+ +
5) ( )
2
'' 2 sin 4 cos y x x x x +
6)
2
'' 4 sin ( 3) cos y x x x x + 7) y = -4sin2x 4xcos2x
8) y = -29sin5x.cos2x 20cos5x.sin2x
Bi 16: Tnh o hm cp n ca cc hm s sau:
a)
1
1
y
x

+
b) y = sinx
S: a)
( )
( )
( )
1
!
1
1
n
n
n
n
y
x
+

+
b)
( )
sin
2
n
y x n
_
+

,
7
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
B. HNH HC
I. CC DNG BI TP THNG GP
Dng 1 : Chng minh hai ng thng a v b vung gc
Phng php 1: Chng minh gc gia hai ng thng a v b bng
0
90 .
Phng php 2:
. 0 a b u v
r r
(
, u v
r r
ln lt l vect ch phng ca a v b).
Phng php 3: Chng minh
( ) a b
hoc
( ) b a
Phng php 4: p dng nh l 3 ng vung gc (
' a b a b
vi b l hnh chiu ca t b ln
mp cha t a).
Dng 2 : Chng minh ng thng d vung gc vi mp (P).
Phng php 1: Chng minh: d a v d b vi a b = M; a,b (P)
Phng php 2: Chng minh d // a, a (P)
Phng php 3: Chng minh: d (Q) (P), d a = (P) (Q).
Phng php 4: Chng minh: d = (Q) (R) v (Q) (P), (R) (P).
Dng 3 : Chng minh hai mp (P) v (Q) vung gc.
Phng php 1: Chng minh (P) a (Q).
Phng php 2: Chng minh (P) // (R) (Q).
Phng php 3: Chng minh (P) // a (Q).
Dng 4 : Tnh gc gia 2 t a v b.
Phng php: - Xc nh t a// a, b// b ( a b = O)
- Khi : (a, b) = (a, b).
Dng 5 : Tnh gc gia t d v mp(P).
Phng php: Gi gc gia t d v mp(P) l
+) Nu d (P) th = 90
0
.
+) Nu d khng vung gc vi (P): - Xc nh hnh chiu d ca d ln mp(P)
- Khi : = (d,d)
Dng 6 : Tnh gc gia hai mp (P) v (Q).
Phng php 1:
- Xc nh a (P), b (Q).
- Tnh gc = (a,b)
Phng php 2: Nu (P) (Q) = d
- Tm (R) d
- Xc nh a = (R) (P)
- Xc nh b = (R) (Q)
- Tnh gc = (a,b).
Dng 7 : Tnh khong cch.
Tnh khong t mt im M n t a:
Phng php:
( , ) d M a MH
(vi H l hnh chiu vung gc ca M trn a).
Tnh khong t mt im A n mp (P):
Phng php: - Tm hnh chiu H ca A ln (P).
8
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
- d
(M, (P))
= AH
Tnh khong gia t v mp (P) song song vi n : d
(, (P))
= d
(M, (P))
(M l im thuc ).
Xc nh on vung gc chung v tnh khong gia 2 t cho nhau a v b:
+) Phng php 1: Nu a b :
- Dng (P) a v (P) b
- Xc nh A = (P) b
- Dng hnh chiu H ca A ln b
- AH l on vung gc chung ca a v b
+) Phng php 2:
- Dng (P) a v (P) // b.
- Dng hnh chiu b ca b ln (P). b // b, b a = H
- Dng t vung gc vi (P) ti H ct t b ti A.
- AH l on vung gc chung ca a v b.
+) Phng php 2:
- Dng t (P) a ti I ct b ti O
- Xc nh hnh chiu b ca b trn (P) (b i qua O).
- K IK b ti K.
- Dng t vung gc vi (P) ti K, ct b ti H.
- K t i qua H v song song vi IK, ct t a ti A.
- AH l on vung gc chung ca a v b.
II. BI TP
Bi 1: Cho hnh chp S.ABC c y ABC l tam gic vung ti B. SA (ABC).
a) Chng minh: BC (SAB).
b) Gi AH l ng cao ca SAB. Chng minh: AH SC.
Bi 2: Cho hnh chp S.ABCD c y ABCD l hnh vung. SA (ABCD). Chng minh rng:
a) BC (SAB).
b) SD DC.
c) SC BD.
Bi 3: Cho t din ABCD c AB=AC, DB=DC. Gi I l trung im ca BC.
a) Chng minh: BC AD.
b) Gi AH l ng cao ca ADI. Chng minh: AH (BCD).
Bi 4: Cho hnh chp S.ABCD c y l hnh vung, tm O v SA = SC = SB = SD =
2 a
.
a) Chng minh SO (ABCD).
b) Gi I, K ln lt l trung im ca AB v BC. Chng minh IKSD
c) Tnh gc gia t SB v mp(ABCD).
Bi 5: Cho t din ABCD c AB CD, BC AD. Gi H l hnh chiu ca A ln mp(BCD). Chng minh:
a) H l trc tm BCD.
b) AC BD.
Bi 6: Cho t din u ABCD. Chng minh rng cc cp cnh i din ca t din vung gc vi nhau tng i
mt.
9
0
BAD 60
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
Bi 7: Cho hnh chp S.ABCD c y l hnh ch nht, tm O v AB = SA = a, BC = 3 a , SA (ABCD).
a) Chng minh cc mt bn ca hnh chp l nhng tam gic vung.
b) Gi I l trung im ca SC. Chng minh IO (ABCD).
c) Tnh gc gia SC v (ABCD).
Bi 8: Cho hnh chp S.ABCD c y l hnh vung, tm O v SA (ABCD) . Gi H, K ln lt l hnh chiu
vung gc ca A ln SB, SD.
a) Chng minh BC (SAB), BD (SAC).
b) Chng minh SC (AHK).
c) Chng minh HK (SAC).
Bi 9: Cho hnh chp S.ABC c y ABC l tam gic vung ti A, SA = AB = AC = a, SA (ABC).
Gi I l trung im BC.
a) Chng minh BC (SAI).
b) Tnh SI.
c) Tnh gc gia (SBC) v (ABC).
Bi 10: Cho hnh chp S.ABC c y ABC l tam gic vung ti B. SA (ABC) v SA = a, AC = 2a.
a) Chng minh rng: (SBC) (SAB).
b) Tnh khong cch t im A n mp(SBC).
c) Tnh gc gia (SBC) v (ABC).
d) Dng v tnh di on vung gc chung ca SA v BC.
BI TP TNG HP
Bi 1: Cho t din OABC c OA , OB , OC i mt vung gc v OA= OB = OC = a.
Gi I l trung im BC; H, K ln lt l hnh chiu ca O ln trn cc ng thng AB v AC.
1. CMR: BC(OAI).
2. CMR: (OAI) (OHK).
3. Tnh khong cch t im O n mp (ABC). S:
a/ 3
5. Tnh csin ca gc gia OA v mp (OHK). S:
cos 6 / 3
6. Tnh tang ca gc gia (OBC) v (ABC). S: tan 2
7. Tm ng vung gc chung ca hai ng thng HK v OI. Tnh khong cch gia hai
ng y. S: a/ 2
Bi 2: Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a,
SA (ABCD)
v
SA a 2
.
1. CMR: Cc mt bn ca hnh chp l cc tam gic vung.
2. CMR: mp (SAC) mp(SBD) .
3. Tnh gc

gia SC v mp (ABCD), gc

gia SC v mp (SAB). S:
0 0
45 , 30
4. Tnh tang ca gc

gia hai mt phng (SBD) v (ABCD). S:

tan 2
5. Tnh khong cch t im A n mt phng (SBC), khong cch t im A n mp (SCD).
S:
a 6/ 3
6. Tm ng vung gc chung ca cc ng thng SC v BD. Tnh khong cch gia hai
ng thng y. S: a/ 2
7. Hy ch ra im I cch u S, A, B, C, D v tnh SI. S: SI a
Bi 3: Cho hnh chp S.ABCD c y ABCD l hnh thoi tm O cnh a, SA SB SD a 3/ 2
v . Gi H l hnh chiu ca S trn AC.
1. CMR: BD
(SAC)
v
SH (ABCD)
.
2. CMR: AD SB .
3. CMR: (SAC) (SBD).
4. Tnh khong cch t S n (ABCD) v SC. S:
SH a 15/ 6
v SC =
a 7/ 2
10
0
ADC 45
Trng THPT Nguyn Trung Trc cng n tp HKII lp 11 - Gio vin: Nguyn Hong Diu
5. Tnh sin ca gc

gia SD v (SAC), csin ca gc

gia SC v (SBD).
S:
sin 3/ 3
v cos 3/ 14 .
6. Tnh khong cch t H n (SBD). S:
a 10/12
7. Tnh gc gia
(SAD)
v (ABCD). S: tan 5
8. Tm ng vung gc chung ca cc ng thng SH v BC. Tnh khong cch gia hai
ng thng y. S:
a 3/ 3
9. Hy ch ra im I cch u S, A, B, D v tnh MI. S:
3 15a/ 20
Bi 4: Cho hnh chp S.ABCD c ABCD l hnh thang vung ti A, AB = BC = a v .
Hai mt bn SAB, SAD cng vung gc vi mt y v SA = a 2.
1. CMR: BC mp(SAB).
2. CMR: CD SC .
3. Tnh gc

gia SC v (ABCD), gc

gia SC v (SAB), gc

gia SD v (SAC).
S:
0 0
45 , 30 , tan 2 / 2
4. Tnh tang ca gc

gia mp(SBC) v mp(ABCD). S: tan 2

5. Tnh khong cch gia SA v BD. S:
2a/ 5
6. Tnh khong cch t A n (SBD). S:
2a/ 7

7. Hy ch ra im M cch u S, A, B, C; im N cch u S, A, C, D.
T tnh MS v NS. S: MS a ,
NS a 6 / 2
Bi 5: Cho hnh lp phng ABCD.ABCD c cch a. Gi O l tm ca t gic ABCD; v M, N ln lt l
trung im ca AB v AD.
1. CMR: BD
(ACC'A')
v AC
(BDC')
.
2. CMR: A'C AB' .
3. CMR: (BDC) (ACCA) v (MNC) (ACCA).
4. Tnh khong cch t C n mp(BDC). S:
a/ 3

5. Tnh khong cch t C n mp(MNC). S:
3a/ 17
6. Tnh tang ca gc gia AC v (MNC). S:
tan 2 2/ 3
7. Tnh tang ca gc gia mp(BDC) v mp(ABCD). S: tan 2
8. Tnh csin ca gc gia (MNC) v (BDC). S: cos 7/ 51
9. Tnh khong cch gia AB v BC. S:
a 3/ 3
11