Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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Bi=)\)j.c*)>%d6kF!#$%!$&'!()!

y =
1
3
x
3
−
5m
2
x
2
− 4mx + 2 (1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";!5<1!

m = −1
*!
=* >?'!'!7@!:";!AB!$C1!A/A!.D9!

x

(x +1)
*!
0; >?'!J1-!.D9!K<3!3$F.!5&!3$L!3$F.!AMC!$&'!()!

f (x) = (x +1).e
x
2
−x
!.D43!7%N3!OP"Q=R*!!!
Bi=)7)j\c*)>%d6kF!>S3$!.SA$!G$T3!

I = (x −1).cos
2
x
2
dx
0
π
∫
*!
Bi=)l)j\c*)>%d6kF))
C; #$%!()!G$UA!

z = 1+ i 3
*!V12.!W!XH<1!XN3J!KHY3J!J1-A*!>?'!3!3JZ[43!XHI3J!3$L!3$F.!7@!

z
n
K&!()!3JZ[43!XHI3J*!
0; #$%!=3!71@'!

Bi=)a)j\c*)>%d6kF!E1,1!$r!G$HI3J!.D?3$!

x
2
−3y
2
+ x + 4y − 2 =
( y −1)
2
+1
x
y
2
−3x
2
− 2x − 2y + 2 = −2.
x
2
+ x
y
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪

+ ab
c
2
+ (a +b)
2
−
8 3(a
2
+ b
2
+ c
2
+ 2)
5
*!
mmm!nLmmm)
)
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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=!
PHÂN TÍCH BÌNH LUẬN VÀ ĐÁP ÁN
Bi=)\)j.c*)>%d6kF!#$%!$&'!()!

y =
1
3
x
3

*!!
"* €hA!(13$!./!J1,1*!
=* >C!AB•!

y ' = x
2
− 5mx − 4m; y ' = 0 ⇔ x
2
− 5mx − 4m = 0
*!
‚@!:";!AB!$C1!A/A!.D9!o$1!5&!A$\!o$1![ƒ!AB!$C1!3J$1r'!G$T3!01r.!!
!

⇔ Δ = 25m
2
+16m > 0 ⇔
m > 0
m <−
16
25
⎡
⎣
⎢
⎢
⎢
⎢
*!
+$1!7B!!.$„%!51P….!AB!

x

+12m) =1
⇔ (x
1
2
− 5mx
1
− 4m +5m(x
1
+ x
2
) +16m)(x
2
2
− 5mx
2
− 4m +5m(x
1
+ x
2
) +16m) = 1
⇔ (5m(x
1
+ x
2
) +16m)
2
= 1 ⇔ (25m
2
+16m)
2

+ 5mx
1
+12m
m
2
7N.!J1-!.D9!3$L!3$F.*!!!!!
€y•!!

A =
m
2
(x
1
2
−5mx
1
− 4m)+ 5m(x
1
+ x
2
) +16m
+
(x
2
2
−5mx
2
− 4m)+ 5m(x
1
+ x

= 2
*!
yFZ!0b3J!u,[!DC!o$1!

m
5m +16
=
5m +16
m
= 1 ⇔ 5m +16 = m ⇔ m = −4(t / m )
*!
‚‡(•!

m = −4
*!
Bi=).)j\c*)>%d6kF)
C; E1,1!0F.!G$HI3J!.D?3$!

1
log
2
(x
2
− x + 2)
≥
1
log
2
(x +1)
*!

(x −
1
2
)
2
+
7
4
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
≥ log
2
7
4
> 0
*!!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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~!
ˆ;!j2Z!

−1< x < 0 ⇒ log

oU%)#;@)#<p&1)#q)m!E1,1!0F.!G$HI3J!.D?3$!

1
log
2
(x
2
− 3x + 4)
>
2
log
2
(x +1)
*!
‚‡(•!

S = (−1;0) ∪ (1;3)
*!!!!
0; €&'!()!Š:u;!K143!.sA!.D43!7%N3!OP"Q=R*!
>C!AB•!

f '(x) = e
x
2
−x
+ (x +1)(2x −1).e
x
2
−x
= (2x

.e
3
4
; f (1) = 2; f (2) = 3e
2
*!
V?!5†[!

max
x∈ −1;2
⎡
⎣
⎢
⎤
⎦
⎥
f (x) = f (2) = 3e
2
; min
x∈ −1;2
⎡
⎣
⎢
⎤
⎦
⎥
f (x) = f (−1) = 0
*!!!!
Bi=)7)j\c*)>%d6kF!>S3$!.SA$!G$T3!


(x −1)cos x dx
0
π
∫
M
! "####### $#######
*!
ˆ;!

K =
1
2
(x −1)dx
0
π
∫
=
1
2
(
x
2
2
− x )
π
0
=
π
2
4

π
∫
= cosx
π
0
= −2
*!
HC#)$=;&()V†[!

K =
π
2
4
−
π
2
−1
*!
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C; #$%!()!G$UA!

z = 1+ i 3
*!V12.!W!XH<1!XN3J!KHY3J!J1-A*!>?'!3!3JZ[43!XHI3J!3$L!3$F.!7@!

z
n
K&!()!3JZ[43!XHI3J*!
0; #$%!=3!71@'!

(n ≥ 2,n ∈ !)

+ i.sin
π
3
)
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
n
= 2
n
(cos
nπ
3
+ i.sin
nπ
3
)
*!
‚@!

z
n
K&!()!3JZ[43!XHI3J!o$1!5&!A$\!o$1!

sin

*!
V†[!

n = 6
K&!J1-!.D9!Ak3!.?'*!!!!!!
0; ˆ;!>C'!J1-A!5Z`3J!7HYA!.N%!.$&3$!.a!'].!7Hq3J!oS3$!AMC!7Hq3J!.D‹3!3]1!.12G!7C!J1-A!7^Z!
5&!"!7\3$!o$`3J!3b'!.D43!7Hq3J!oS3$*!
ˆ;!#B!.F.!A,!3!7Hq3J!oS3$*!V<1!'Œ1!7Hq3J!oS3$!AB!:=3P=;!7\3$!A‹3!KN1!A•3J!5<1!7Hq3J!oS3$!7B!
7@!.N%!.$&3$!'].!.C'!J1-A!5Z`3J*!
V†[!AB!.F.!A,!

n.(2n − 2) = 2(n
2
− n)
.C'!J1-A!5Z`3J*!
>$„%!J1,!.$12.!.C!AB•!

2(n
2
−n) = 180 ⇔
n =10(t / m)
n = −9(l )
⎡
⎣
⎢
⎢
*!!
V†[!

n =10

2
AM .BC =
a
2
3
4
*!
eZ[!DC•!

V
S .ABC
=
1
3
SM .S
ABC
=
1
3
.
a
2
.
a
2
3
4
=
a
3

2
+ BC
2
)− SC
2
2
=
a 10
4
*!
eZ[!DC!

cos BNE
!
=
BN
2
+ NE
2
− BE
2
2BN .NE
=
a
2
2
+
a
2
4

a
2
;0),S (0;0;
a
2
),N (
a 3
4
;0;
a
4
)
*!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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"!
#$%!&'!

cos(BN ; AC )
!
=
BN
" #""
.AC
" #""
BN
" #""
. AC

+ MB
2
= 2MI
2
+
AB
2
2
(!
fU!1g%!

MA
2
+ MB
2
+.X!+.Y4! 0!V]!+.X!+.Y4!F!80h$!+J%!4di+,!8di+,!120!V!^J!.U+.!6.0S$!
1$/+,!,_6!6R'!]!4&W+!@TE(!
ZE!jde+,!4.M+,!k!80!l$'!]!1J!1$/+,!,_6!120!@TE!+.g+!

n
P
!"!
= (1;1;1)
^J>!1m6!4i!6.n!L.di+,!+W+!6_!
L4!^J!

d :
x = 4 + t
y = 1+ t
z = 3+ t

⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
(!
fg%!80=>!6p+!4U>!^J!

M −
4
3
;−
13
3
;−
7
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟

BD ⊥ AC
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ JK ⊥ MI (1);
MK //AD
IJ ⊥ AD
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ IJ ⊥ MK (2)
(!!
)Q!@AE!1J!@IE!v$%!&'!]!^J!4&•6!4s>!4'>!,0P6!Vu~(!
fU!1g%!

IK ⊥ MJ
IK //NJ
⎧
⎨
⎪
⎪
⎩

x = 5
y =
7
2
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇒ J (5;
7
2
)
(!
fU!u!^J!4&$+,!80=>!Gt!+W+!G@"BAE(!
[\0!q@'ByE!120!'€`!4'!6_!

BC = 2NC =
2BN
5
= 2 5
(!!!
)'!6_!.3!L.di+,!4&U+.a!

(a − 5)
2

⎩
⎪
⎪
⇔
a = 7,b = 5(t / m)
a = 3,b = 5(l)
⎡
⎣
⎢
⎢
(!
fU!1g%!

C (7;5)
F!k*!t!^J!4&$+,!80=>!qr!+W+!r@`BHE!1J!

DC
! "!!
= AB
! "!!
⇒ A(1;3)
(!
fg%!4*7!89!y}+!8n+.!6p+!4U>!^J

A(1;3),B(5;1),C (7;5),D(3;7)
(!
BH=)O)IJK*)>%L6MF![0O0!.3!L.di+,!4&U+.!

x
2

j0h$!-03+a!

x ≠ 0; y ≠ 0
(!
•3!L.di+,!4&U+.!4di+,!8di+,!120a!

x
3
−3xy
2
+ 4xy = y
2
− x
2
+ 2(x − y +1) (1)
y
3
−3x
2
y − 2xy = 2( y
2
− x
2
− x − y) (2)
⎧
⎨
⎪
⎪
⎩
⎪

⎣
⎢
⎢
⎢
⎢
⎢
⇔
x = −1, y = 0
x =1, y =1
z = −1, y =1
⎡
⎣
⎢
⎢
⎢
⎢
⎢
(!
PC#)$=;&()fg%!.3!L.di+,!4&U+.!6_!.'0!+,.03>!^J!

(x; y) = (1;1);(−1;1)
(!!!
B4Q)3F!)'!6_a!

z
2
= x
2
− y
2

x
3
−3xy
2
− x −1 = y
2
+ 2xy − x
2
y
3
−3x
2
y + y +1= x
2
+ 2xy − y
2
⎧
⎨
⎪
⎪
⎩
⎪
⎪
(!
BH=)+)IJK*)>%L6MF!q.*!'FyF6!^J!6P6!v}!4.•6!kdi+,!4.*O!>„+!

ab + bc + ca =1
(!)U>!,0P!4&{!^2+!+.Y4!
6R'!y0=$!4.c6!


2
+ 2)
5
(!
Lời$giải:$
#w!k5+,!yY4!8M+,!4.c6!?V!…[V!4'!6_a!
!

a
2
+ bc
a
2
+ (b + c )
2
≤
a
2
+
(b + c )
2
4
a
2
+ (b + c )
2
= 1−
3
4
.

+ (b + c )
2
∑∑
(!!
#w!k5+,!yY4!8M+,!4.c6!q'$6.%!…#6.†'&<!4'!6_a!
!

(b + c )
2
a
2
+ (b + c )
2
≥
4(a + b + c)
2
(a
2
+ (b + c )
2
)
∑
=
4(a + b + c)
2
(a + b + c)
2
+ 2(a
2
+ b

)
∑
=
6(a
2
+ b
2
+ c
2
)
(a + b + c )
2
+ 2(a
2
+ b
2
+ c
2
)
≤
6(a
2
+ b
2
+ c
2
)
(a + b + c )
2
+

2
(a + b + c)
2
−
8 3(a
2
+ b
2
+ c
2
+ 2)
5
=
18
5
(1−
2
(a + b + c)
2
)−
8 3
5
(a + b + c)
= f (t ) =
18
5
−
36
5t
2

= −
18
5
(!
fg%!,0P!4&{!^2+!+.Y4!6R'!T!yŠ+,!

−
18
5
(!rY$!yŠ+,!8K4!470!

a = b = c =
1
3
(!!!!!
!
!
!!
!!