!"!#$%&'()không kể thời gian phát đề)
*+,-./01
23)4!567
( )
y x m x m= − − + −
( )
m
C
( )
C
!
m =
"
#m$!%&!'()*+"
23)4!567
,-./)%0
( ) ( ) ( )
tan x sin x tan x .− + = +
,1-./)%0%2-3
( )
!
))E>.F)G)
AC
MN
?)
!
a
"
238)4!567x, y :H
!x xy y .+ + =
60)%I@J:K
@$+L3
! !
M 5M x y xy= + −
"
*+,-9:01
Dành cho thí sinh thi theo chương trình chuẩn
238*#)4!567
6%)1N(Oxy, 0+O);<=$
( )
P!A −
-./)%0
.F)G)
( )
3 4 QBD x y− + =
"60N('R:&0+O)"
6%) O))STAU$
( )
! 9 Qz z z z− + + − + =
Dành cho thí sinh thi theo chương trình nâng cao
238*;)4!567
W-./)%0.F)%R
( )
C
8I+(
( )
3! Qx y∆ + − =
-TY
K.F)G)
( )
3 4 Qd x y+ + =
( )
3 9 Qd x y− + =
1
W-./)%0V-G)
( )
α
X+$
( )
QPQPM
P
( )
QPPQN
x
y'
x
=
= ⇔
= ±
?@
•
( ) ( )
Q
CD CT
y y ; y y .= = = = −
?@
• <)3
x
−∞
] Q
+∞
y'
−
Q
+
Q
−
Q
=
′
= ⇔
= −
!% m > 1
?@
• WK b!$%:3
( ) ( )
( )
( )
( )
Q Q 4 Q 4A ; m ,B m ; m m ,B m ; m m .− − − + − − − − + −
63
( ) ( )
( )
c
M
AB AC m m
BC m
= = − + −
= −
?@
• ^*+ 1);<*+:
?@
• \K*+ 1!%+A%
!
!
m = +
3
?@
23 4!56
#7 ,-./)%0
( ) ( ) ( )
tan x sin x tan x .− + = +
• ^*+ 13
d
d
x k ,k≠ + ∈Z
• <e-./)%0*>&)
( ) ( )
Q
tan x
sin x cos x c x
• W:&1>.K>&)3
( ) ( )
( ) ( )
c
5
x x xy y
x x xy y
+ + + + =
+ + + =
?@
• ^V
u x= +
v x xy y= + +
P1%Z3
c
!
5
u v
⇔ ⇔ = − − − −
+ =
+ + =
?@
• W2A1)1.%"
?@
23 677-83
9
!
x
I dx
x x
−
=
+
∫
• ^e
c
t x=
( )
! !
!
= − + + −
= − + −
÷
WK
!
dt
J
t
=
+
∫
?@
• ^$7JV
t tan x.=
!
!
J dt
π
π
π π π
= = − =
∫
?@
MK⇒ =
"
?@
• W0
A B AB MK A B⊥ ⇒ ⊥
( )
CB ABB A CB MK⊥ ⇒ ⊥
"
• 6f+A%
( ) ( )
( )
( )
MK A BC MK d MN , A BC d MN,AC⊥ ⇒ = =
• _
! ! !
a x a a
MK x= ⇒ = ⇒ =
"W2ADIH
!
a
BM =
?@
238 Tìm giá trị nhỏ nhất, lớn nhất của biểu thức:
! !
∈ −
J.C3
( )
( )
!Q
!
4
min f t min f ; f
= − −
÷
÷
( )
( )
!Q
T T !
4
f t f ; f
= − + −
÷
4 I t ;t⇒ + +
"
?@
• \j>a)*+ 1
( )
BD
AI u⊥
uur uuuur
+A%
( )
!
t I ; C ; .
= − ⇒ − ⇒ − −
÷
?@
• W0
( ) ( )
4 B BD B t ;t∈ ⇒ + +
"=
QAB CB AB.CB⊥ ⇒ =
uuur uuur uuur uuur
• l./)%0.F)G)B0:3
( )
!
3
M
x y z
d
− + −
′
= =
− − −
A
( )
!
3
x y z
d
− + −
′
= =
?@
238*; Chương trình cơ bản
• ^V
t z z= −
0-H%Z3
238*# Chương trình nâng cao
#7
• ^.
( )
∆
*>&)
( )
3 P
!
x t
t
y t
= +
∆ ∈
= − −
R
"
• ,N
( ) ( )
P ! I t t+ − − ∈ ∆
m:B:.C:8 7.F)
%R"
?@
• 6f - TY +A %
( )
( )
= −
?@
• 6f>n-"
?@
;7
• ,N
( )
3 ;T QBy Cz D
α
+ + + =
Q
C B
C D B D
D B
=
⇒ + = + = ⇒
= −
?@
• D-
( )
α
&K-
( )
− − = ⇒ =
+A%J="
• :+23-H *"
?A@
238*; Chương trình nâng cao
• 63
( )
Q
"""
n
n n n
n n n
x C x C x C
+
+ +
+ + +
− = − + −
P
( )
Q
"""
n
n n
n n n
x C C x C x
+
− = − + + −
[1L
n
x
+
%) %$h?)QJ%) %$h:3
?A@
4
( ) ( ) ( ) ( )
Q
"""
n
n n n n
C C C C
+
+ + + +
− + − −
• W2A ) @ 1 .C3
( ) ( ) ( ) ( )
Q
""" Q
n
n n n n
C C C C
+
π
+=
+
+ x
xx
x
x
* ,@-./)q3
!4 4 x x x+ < − + +
23)?4!567" 677-83
h 4
xdx
x x x
π
π
−
− +
∫
238)?4!567"
p
:r)%a)*+
sss" CBAABC
3
4 Qx y+ − =
"W
-./)q.F)G)h>X+$DhQP&Kh>
Jh>
()8&)$
h>
Jh>
"
* $;hP4PJ<hQPPJhPP"6
p
N($=+(.F)G);<
(>&G)=I@"
238*# (1,0 điểm). ,-./)q+%2--LhU
t!Utc
tUhU
t!Utc]!U
oQ
c
1
*CDEFG$"(H$2$"E#D
238*;(2,0 điểm)
?Y:+M$
(!WLZ!*Thời gian làm bài : 150 phút không kể thời
gian giao đề
23 K![3$" !56
i
ND\J(\];!^$(!Q$_`_a4U(V)7Eb#`6\Z)?4!567
]62-T3mxyz]\3
( )
s Q
y x
x
−
= < ∀ ≠
−
"[)%{ )
( )
P−∞
( )
P+∞
Q"4
]
( ) ( )
: P :
x x
y y x
]^3[N"|+B+8J7TL)X+)$
.F)12"6$1Y))$K%a&("
Q"4
6
p
&($<J} JQ
6
h 3
C y
x
= +
−
P,N
h P J h P J
B b C c
b c
+ +
− −
Kh~~"
,N[J:B:.C:
p
+<J:%aSTJ
QJ4
9
·
·
!
b
b
c
c
c
b
− = +
= −
−
⇔
=
+ = −
−
"W2A
h PJ h!P!B C−
"
QJ4
h J •
!
x m
x x m
x x m n
n
x
x x n
π
π
π
π
π
π π
π
π π
= +
= + +
= + ⇔ ⇔ ∈
= +
= − − +
Q"4
,@-./)q}" JQ
<l6./)./)3
!4 4 4 h4 h !4
!4
x x x x x x x
x x
+ − + < − ⇔ < − ⇔ < − + + +
+ + +
Q"4
_+T
4
≤
O)Ir<l6
Q"4
_+Tb€43[
h4 h !4 y x x x= − + + +
KTb€4A
k
o
4h !4 h4 h
!4
∫
"^V
dt
t x dx
t
= ⇒ =
+
"6
: !
4 ! 4
t dt dt
I
t t t t
− −
= = + −
− + − +
∫ ∫
Q"4
67
= + −
"
QJ4
iW JQ
[
p
W
M
Kẻ
s h s sBD AB D A B
Q
cQsJhsJsh == BCBDBCAB
Q
cQs= DBC
hoặc
"Qs
Q
=DBC
QJ4
Nếu
Q
cQs=DBC
. Vì lăng trụ đều nên
s h s s sJBB A B C
áp dụng định lý Pitago và định lý cosin ta có
s
+== mBCBD
2 1 2 1
2 2 0
1 1
x x
m
x x
ổ + ử ổ + ử
ữ ữ
ỗ ỗ
- + =
ữ ữ
ỗ ỗ
ữ ữ
ỗ
ữ
ỗ
ố ứ
ố ứ
+ +
"
QJ4
^V
2
2 1
1
x
t
x
+
=
! ! Qh
9 9 4
! Qh
h 9
x y
x y x y
x y
+ =
+ +
=
=
+ +
QJ4
l6.F)B
p
X+DhQP))K
J
.F)G)r
! ! Qx y+ =
! Qx y + =
QJ4
6
p
&($=} JQQ
6
ob]]ct]5t5oQ~ob
c
a =
"6N($
4 5
P P
c c c
D
ữ
Q"4
Wii
,-./)q%2--L JQQ
6@AUoQ O):)1-./)q"U
V
2
3 6z z
t
z
+ +
=
J
=nK-./)q3
t]!oQoVo]!"
QJ4
y x
x y
= =
+ + − − =
⇔
= − = −
− − =
"W
p
;
(>./).C;hPQJ<h]!P]"
QJ4
W
p
·
Q
5QABC =
;:.F) 7.F)uJL:$TL)K$;X+8i
.F)u"68ih]PJ+A%h]P"
QJ4
g
6(Dh4KEJE4!56?
∆
:.F)G)X+$i+O))KhlJ
p
_
Q
:)$
∆
hl"^.F)G)
∆
‚/'-./):
( )
PP
P
n = −
r
X+i-./)q:
( )
!
x t
y t t
z t
= +
= − + ∈
= −
uuuur uuur
\+A%D
Q
hQP]!P
QJ4
Wii
,-./)q%%2--L""" JQQ
"U
]U
!
t
2
2
z
tUtoQ⇔hU
t]hU
!
]Ut
2
2
z
oQ"
QJ4
U
J.C3hU
iw = -
tl./)q3U]
1
z
o
1
2
t
3
2
)1U
otPU
o]
1
2
h]t
l./)q3U]
1
z
o
1
2
]
3
2
)U
!
o]
x
x
x
+
+ − = +
"
"60m$1-./)%03
! !
! ! Q
! Q
x y y x
x x y y m
− + − − =
+ − − − + =
)1"
Câu 3 (2.0 điểm):"6%) O))K1N(OxyzJV-G)hP.F)
G)hd:B:.C-./)%03
hl3x−y−z−oQP hd3
x y z+ −
= =
−
"W-./)%0VB+8+(.F)G)hdJV-G)hP( )
-%:hP3y
ox"
"601&)Lx
M
%) %$_‚ƒ3
x
x
− −
÷
−−−−−−−−−−−−−Q−−−−−−−−−−−−−
Cán bộ coi thi không giải thích gì thêm.
Họ và tên thí sinh: SBD:
8+ _(>+)
^$
"moJ>&)3yox
!
−!x
t
t6#^3
t\3yko!x
LËp BBT:
0.25
§å thÞ:
0.25
€"63yko!x
−cmxoQ⇔
Q
x
x m
=
=
^$&$+0m≠Q"
Q"4
Q
x
t†
−†
−
t
t
Q
Q
y’
=
Q"4
,%3
m = ±
PmoQ Q"4
C-K*+ 13
m = ±
€"^ 3
x k
π
≠
Q"4
l./)%0H./)./)K3
( )
! !
π
= − + π
= −
⇔
π
=
= + π
Q"4
‡3\K*+ 1-./)%0)13
c
x k
π π
= +
Pk∈Z Q"4
€"
! !
! ! Qh
! Qh
Q"4
^Vtoxt⇒t∈ˆQP‰Ph⇔t
!
−!t
oy
!
−!y
"
Q"4
[fhuou
!
−!u
)%&ˆQP‰3
h⇔yoy⇔yoxt⇒h⇔
Qx x m− − + =
Q"4
!
^V
v x= −
⇒v∈ˆQP‰⇒h⇔v
tv−om"
[ghvov
tv−&
− + − − − − +
∆ = = =
⇔
!
9
!
t
t
=
= −
Q"4
⇒8VB+3
M 9 9
P P P P
! ! ! ! ! 9
vµ I I
− − −
÷ ÷
W0V-G)hPvVB+‚.F)%R 7?)VB+
7:Ro4"
Q"4
!
!" c
− − −
α = =
⇒,)EV-G)hQV-G)hQB0:
c !
5 !
α = − =
Q"4
,jhQX+h∆>&)3mhxtyttnhxtz−oQhm
tn
bQ
⇔hmtnxtmytnztm−noQ
W2A))EhPhQ:3
Š ! Š !
!
!" 4
m
m n mn
α = =
+ +
Q"4
⇔m
4 ! 4
x
x
π
π − − =
÷
Q"4
€"63
[ ]
h h h 5
xy yz zx
xy yz zx
+ + + + + + + ≥
÷
+ + +
Q"4
5 5
!
!
P
xy yz zx
x y z
⇔ ≥ ≥
oACh
Q"4
6hh3CoAVCo−A"
WKCo−A⇒AoBoQh:&
Q"4
WKCoA⇒
!
A
B = ±
⇒^.F)G)H-./)%03
!
Q Q
!
!
A
Ax y A x y± + = ⇔ ± + =
Q"4
W2A-+AB03
!
Q
!
x y± + =
Q"4
4
8
63
Q Q
h h
h
i
k k
k i
k k i k k i k i i
k k
k i k i
k
k k i k i
k
k i
C C x C C x x
x
C C x
−
− − − −
= = = =
− −
= =
= − = −
÷
= −
∑ ∑ ∑∑
∑∑
23"h4,0 điểm
" ,1-./)%03
M
4"
x x y x y y
x y
− = +
− =
hx, y∈m
" ,-./)%03
h
x x x
π
+ = + −
"hx∈m
23"h2,0 điểm
-./)%03
:)h Q :)h x x m x+ + = +
hKm:h
60m$-./)%0h)1-81"
238"h2,0 điểm
677-83
&)%I@"
238"(2,0 điểm)
0 - S"ABCD A ABCD : 0 +O) & a, SA +O)
)KA",)EV-G)hSBChSCD?)cQ
Q
"
67‚a$7 -S"ABCD"
238"h1,0 điểm
>./)a, b, cIHab + bc + cao!"
L)%?)3
! ! !
!
! ! !
a b c
b c a
+ + ≥
+ + +
"
hCán bộ coi thi không giải thích gì thêm
k_`(Q$(O\!$}}}}}}}}}}}}}}}"1,}}}}}}}}
l,m.-
23 FG$"&J&W^(n3N !56
*
)4!567
"6Ako3x
2
+ 6x + m QJ4
|./)./)K-./)%03x
2
=
+ =
QJ4
,1%.Cmo]Q4 QJ4
*
)4!567
"t[($+)h>:)1-./)
%0
T
!
t!T
tTto⇔ThT
t!TtoQ
QJ4
6f0.C~
5
≠Q0>vh&$-8
1;hQPJ<J"
QJ4
t<hT
"
o]
QJ4
⇔
Œ5toQ
QJ4
⇔
5 c4
h€
M
5 c4
h€
M
−
=
+
=
QJ4
*
)4!567
"^*+ 1TJA•Q QJ4
− −
⇔
− =
= ≠
−
hŽ⇔
!
ŒM
tt!oQ
⇔oPo]
Po
!
"^+*+ 1.Co
!
6f0.ChTPAoh5P"
h[\$)?)-./)--V
.C X+Y)n.C$
QJ4
− =
QJ4
L).C-./)%0!TŒToO)1
‡3To
k
π
π
− +
QJ4
)4!567
!"l6⇔
Q h ! c hŽŽ
x x
x x m x m x x
> − > −
⇔
+ + = + = − +
|⇔hŽŽ)1-81HTb]
Q
xdx
x x
π
+
∫
"
QJ4
^Vo
T
>o
dx
x x
x
+ ⇒ = + ⇒
QJ4
^e23ToQ⇒o
To
!
π
⇒ =
QJ4
io
⇔
]Q]tcoQh
]Mo Q Mh
− +
o O):)11%"
QJ4
h⇔o
4]M
]
"6h0.CoQVo
QJ4
WKoQ+A%o"
WKo+A%oc"
QJ4
8*
)4!567
",Ni:%+)$;<⇒ihPP
tD;
tD<
oDi
ti;
ti<
A
S
M
,ND:0++O))<:\"L)
.C)=D<oQ
Q
∆=D<8&D
67.C3=D
o
!
QJ4
∆\=+O)&==D:.F)
o t
=D =\ =
\+A%=\o
"6);\=+O)&;+A%\;o"
QJ4
W2A$7\";<=?)
!
!
QJ4
b c c a
+ +
+ + ≥
+ +
!
4
h h M
a a b c
b c c a
− −
⇒ ≥
+ +
h
QJ4
[./)L).C3
!
4
h h M
b b c a
c a a b
− −
≥
+ +
hJ
!
4
h h M
c c a b
Câu II. (2.0 điểm)
",-./)%0
cTtT] ! T oTt !c c
",1-./)%0
x x
y
y y x y
+ − =
− − = −
Câu III. (1.0 điểm)
677-8
!
Q
h
x
x x dx
x
SA"
"0:2--./);<=";k<kk=k
&?)",ND:%+)$&;=J
_:
80+O)k=k="67 7VB+X+$<JkJDJ_"
Câu VIIa. (1.0 điểm)
,@-./)%0
!
!
:) h :) h
Q
4 c
x x
x x
+ − +
>
− −
B. Theo chương trình chuẩn
Câu VIb. (2.0 điểm)
"$;h]PQJ<hP.F)G)h>3T]A]oQ"‡2--./)%0.F)%R
X+
$;J<-TYK.F)G)h>"
"6%) O))K1%a&(STAU$;hPQPJ<hPPV-G)
h‹3
TtAt!Ut!oQ"‡2--./)%0V-G)hlX+;J<+O))Kh‹"
Câu VIIb. (1.0 điểm)
,-./)%0
x x
f x f x
→+∞ →−∞
= =
Ao:12))
: h J:
x x
f x
+ −
→ →
= +∞ = −∞
To:12L)
Ako
Q
h x
− <
−
Q"4
<)
1
+
∞
-
∞
1
- -
y
Q
PA
Q
+(h-+AK& )f8
TL)-+A::K@"
l./)%0-+A&D>&)3
Q
Q
Q Q
h
h
x
y x x
x x
= − − +
− −
Q
Q Q
Q
h h
x
x y
x x
⇔ − − + =
− −
t t t
t t
− + +
+ +
Q"4
gkhoQ o
<)
f)
>hiP:K@
' oA
Q
Q
Q
Q
x
x
x
=
− = ⇔
=
Q"4
tWKT
Q
oQ-+A:Ao]T
tWKT
⇔
Q"4
9
x k
k
x
k
x
π
π
π π
π π
= +
⇔ = − +
= +
Q
Q
u u v
v v u
+ − − =
+ − − =
Q
! 9 ! 9
J
9 9
u v
u v
u v
u v
v v u
u u
v v
= =
6f)11
h]P]JhPJh
! 9
P
9
−
−
Jh
! 9
P
9
+
+
Q"4
8+iii"
h"Q
!
Q Q
∫
Vo
x
7.Ci
o
Q
h h
dt
t
π π
− = − = −
+
∫
Q"4
6fioi
ti
o]€!h]t
π
−
Q"4
8+iW"
h h h
M
x y z− − − ≤
Q"4
2A;
T
o
!
M
x y z⇔ = = =
Q"4
O
C
B
A
D
S
H
B'
Y
X
Z
N
D'
C'
A'
C
D
A
x
SH
SH SC SA
x
= + ⇒ =
+
W2AWo
! h
c
x x d−
Q"4
8+
Wi"
h"Q
"
h"Q
,N;:)$>
>
;h!PQ
,N<:)$>
M Q
A
A D
B C D
B
A C D
C
B C D
D
= −
+ + =
+ + + =
= −
⇔
+ + + =
= −
+ + + =