đề thi thử thpt quốc gia môn toán,đề số 12

Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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DW&()L"I&X)/Y)Z[)\.]^*)
V1US)#4%)()*\]*7].*\^)
L4_%)1%E&)$U6)`U%()\a*)@4b#c)24W&1)2d)#4_%)1%E&)1%E")>K)
e%f&)4g)>0&1)23)24"I)489)Q)!"#$%&'()*+,-) ).*.)Q)B4%)#%C#()hhhF6E#4$%&2:FG&))
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

1
,x
2
(C%!A$%!

(x
1
2
+ 5mx
2
+12m)(x
2
2
+ 5mx
1
+12m) =1
*!!
Bi=).)j\c*)>%d6kF)
C; E1,1!0F.!G$HI3J!.D?3$!

1
log
2
(x
2
− x + 2)
≥
1
log
2

(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@'!

(n ≥ 2,n ∈ !)
K&!A-A!7\3$!AMC!'].!7C!J1-A!7^Z*!_12.!()!.C'!J1-A!5Z`3J!.N%!
.$&3$!.a!=3!71@'!7B!0b3J!"cd*!>?'!3*!!!
Bi=)^)j\c*)>%d6kF)#$%!$?3$!A$BG!e*f_#!AB!7-[!f_#!K&!.C'!J1-A!7^Z!AN3$!Cg!

SA = a
*!Eh1!igj!
Kk3!KHY.!K&!.DZ3J!71@'!A-A!AN3$!_#!5&!ef*!E1,!(l!ei!5Z`3J!JBA!5<1!'m.!G$n3J!:f_#;*!>S3$!
.$@!.SA$!o$)1!A$BG!e*f_#!5&!A`(13!JBA!J1pC!$C1!7Hq3J!.$n3J!_j!5&!f#*!!!
Bi=)-)j\c*)>%d6kF)>D%3J!o$`3J!J1C3!5<1!$r!.DsA!.%N!7]!tu[W!A$%!$C1!71@'!f:"QdQv;g!_:wQ=Q=;!5&!
'm.!G$n3J!

(P ) : x + y + z + 8 = 0
*!>S3$!o$%,3J!A-A$!.a!.DZ3J!71@'!AMC!7%N3!.$n3J!f_!723!'m.!
G$n3J!:x;*!>?'!.%N!7]!71@'!i!.D43!:x;!7@!

MA
2
+ MB
2
3$L!3$F.*!!!
Bi=),)j\c*)>%d6kF)>D%3J!'m.!G$n3J!5<1!.DsA!.%N!7]!tu[!A$%!$?3$!5Z`3J!f_#y!.T'!z*!Eh1!ig!
jg{!Kk3!KHY.!K&!.DZ3J!71@'!A-A!7%N3!.$n3J!fzg!#yg_j*!E1,!(l!G$HI3J!.D?3$!7Hq3J!.$n3J!i{!K&!

y −
7
2
= 0
5&!j:|Q};*!>?'!.%N!7]!A-A!7\3$!$?3$!5Z`3J!f_#y!012.!7\3$!#!AB!$%&3$!7]!K<3!$I3!~*!!

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
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪

⎪
⎪
⎪
(x, y ∈ !)
*!
Bi=)+)j\c*)>%d6kF!#$%!Cg0gA!K&!A-A!()!.$/A!XHI3J!.$%,!'•3!

ab + bc + ca =1
*!>?'!J1-!.D9!K<3!3$F.!
AMC!01@Z!.$UA!

P =
a
2
+ bc
a
2
+ (b + c )
2
+
b
2
+ ca
b
2
+ (c + a)
2
+
c
2

+ 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

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

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

x
1
,x
2
(C%!A$%!

(x
1
2
+ 5mx
2
+12m)(x
2
2
+ 5mx
1
+12m) =1

*!!
"* €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

1
+ x
2
= 5m
5&!

x
1
2
− 5mx
1
− 4m = 0;x
2
2
− 5mx
2
− 4m = 0
*!!
V?!5†[!!

(x
1
2
+ 5mx
2
+12m)(x
2
2
+ 5mx
1

+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

= 1 ⇔ 25m
2
+16m =1(do25m
2
+16m > 0)
⇔ m =
−8± 89
25
*!!!!
oU%)#;@)#<p&1)#q)m)>?'!'!7@!:";!AB!$C1!A/A!.D9!

x
1
,x
2
(C%!A$%!

A =
m
2
x
1
2
+ 5mx
2
+12m
+
x
2
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
) +16m
m
2
=
m
5(x
1
+ x
2
) +16
+
5(x
1
+ x
2
) +16
m
=
m
5m +16
+
5m +16
m
≥ 2
m
5m +16
.
5m +16
m

= 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)
*!

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*!!!
C; ‚1^Z!o1r3•!

x +1> 0
log
2
(x +1) ≠ 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇔ −1< x ≠ 0
*!
>C!AB•!

log
2
(x
2
− x + 2) = log
2

(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

2
(x +1) < 0
*!_F.!G$HI3J!.D?3$!KZ`3!7‰3J*!
ˆ;!j2Z!

x > 0 ⇒ log
2
(x +1) > 0
*!!!
_F.!G$HI3J!.D?3$!.HI3J!7HI3J!5<1•!
!

log
2
(x
2
− x + 2) ≤ log
2
(x +1) ⇔ x
2
− x + 2 ≤ x +1
⇔ x
2
− 2x +1≤ 0 ⇔ x =1
*!
V†[!.†G!3J$1r'!AMC!0F.!G$HI3J!.D?3$!K&!

S = (−1;0) ∪ 1
{ }
*!!

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

2
+ x).e
x
2
−x
*!
!

f '(x) = 0 ⇔ 2x
2
+ x = 0 ⇔
x = 0
x = −
1
2
⎡
⎣
⎢
⎢
⎢
⎢
*!
ˆ;!>S3$!7HYA•!

f (−1) = 0; f (−
1
2
) =
1
2

.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!

I = (x −1).cos
2
x
2
dx
0
π
∫
*!
>C!AB•!!!

I =
1
2
(x −1)(1+ cos x )dx
0
π
∫
=
1
2
(x −1)dx
0
π
∫
K
! "##### $#####
+
1
2

(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
2
*!
ˆ;!!

u = x −1
dv = cosxdx
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒
du = dx
v = sin x
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ M = (x −1)sin x
π
0
− sin x dx
0

π
∫
= cosx
π
0
= −2
*!
HC#)$=;&()V†[!

K =
π
2
4
−
π
2
−1
*!
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@'!

(n ≥ 2,n ∈ !)

K&!A-A!7\3$!AMC!'].!7C!J1-A!7^Z*!_12.!()!.C'!J1-A!5Z`3J!.N%!
.$&3$!.a!=3!71@'!7B!0b3J!"cd*!>?'!3*!!!
C; >C!AB•!

z = 2(
1
2
+
3
2
.i ) = 2(cos
π
3
+ i.sin
π
3
)
*!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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v!
V?!5†[!

z
n
= 2(cos
π
3

+ 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

nπ
3
= 0
2
n
.cos
nπ
3
∈ ! *
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⇔
nπ
3
= k2π ⇔ n = 6k,k ∈ "
*!
V?!3!3JZ[43!XHI3J!3$L!3$F.!343!

k = 1 ⇒ n = 6

*!
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

K&!J1-!.D9!Ak3!.?'*!!
Bi=)^)j\c*)>%d6kF)#$%!$?3$!A$BG!e*f_#!AB!7-[!f_#!K&!.C'!J1-A!7^Z!AN3$!Cg!

SA = a
*!Eh1!igj!
Kk3!KHY.!K&!.DZ3J!71@'!A-A!AN3$!_#!5&!ef*!E1,!(l!ei!5Z`3J!JBA!5<1!'m.!G$n3J!:f_#;*!>S3$!
.$@!.SA$!o$)1!A$BG!e*f_#!5&!A`(13!JBA!J1pC!$C1!7Hq3J!.$n3J!_j!5&!f#*!!!
!
ˆ;!>C'!J1-A!5Z`3J!efi!AB!

SM = SA
2
− AM
2
= a
2
−
3a
2
4
=
a
2
*!
ˆ;!

S
ABC
=
1

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

3
24
!:75 ;*!
ˆ;!>S3$!JBA!J1pC!_j!5&!f#•!
Eh1!Ž!K&!.DZ3J!71@'!AN3$!e#*!>C!AB!jŽ‡‡f#!343!JBA!J1pC!f#!5&!
_j!0b3J!JBA!J1pC!jŽ!5&!_j*!
>C'!J1-A!_jŽ!AB•!
!

NE =
AC
2
=
a
2
;BN =
2(AB
2
+ SB
2
)− SA
2
2
=
a 2
2
Q

BE =
2(SB

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

−
5a
2
8
2.
a
2
.
a 2
2
=
2
8
*!V?!5†[!

cos(BN ; AC )
!
=
2
8
*!!!
BI94).(!‚m.!$r!.DsA!J8'!i:dQdQd;g!

A(
a 3
2
;0;0),B (0;−
a
2
;0),C (0;

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!
!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&!
"!
#$%!&'!

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

" #""
=
2
8
(!!!!!!
BH=)-)IJK*)>%L6MF))&*+,! /+,!,0'+!120!.3!4&56!4*7!89!:;%<!6.*!.'0!80=>!?@ABCBDEF!G@HBIBIE!1J!
>K4!L.M+,!

(P ) : x + y + z + 8 = 0
(!)N+.! *O+,!6P6.!4Q!4&$+,!80=>!6R'!8*7+!4.M+,!?G!8S+!>K4!
L.M+,!@TE(!)U>!4*7!89!80=>!V!4&W+!@TE!8=!

MA
2
+ MB
2
+.X!+.Y4(!!!
ZE![\0!]!^J!4&$+,!80=>!6R'!?G!4'!6_!]@DBAB`E(!)'!6_a!

d (I ;(P )) =
4 +1+ 3+ 8
1+1+1
=
16 3
3
(!!
ZE!).b*!6/+,!4.c6!8de+,!4&$+,!4$%S+!4'!6_!

MA
2

+ 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

⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
(!).'%!;F%F<!4Q!L4!6R'!k!1J*!L.di+,!4&U+.!6R'!@TE!4'!8do6a!
!

4 + t +1+ t + 3+ t + 8 = 0 ⇔ t = −
16
3
⇒ M −
4
3
;−
13
3
;−
7
3
⎛
⎝
⎜
⎜

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

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

⎟
⎟
(!!
BH=),)IJK*)>%L6MF))&*+,!>K4!L.M+,!120!4&56!4*7!89!:;%!6.*!.U+.!1$/+,!?Gqr!4s>!](![\0!VF!
tFu!^p+!^do4!^J!4&$+,!80=>!6P6!8*7+!4.M+,!?]F!qrFGt(![0O!vw!L.di+,!4&U+.!8de+,!4.M+,!Vu!^J!

y −
7
2
= 0
1J!t@"BxE(!)U>!4*7!89!6P6!8n+.!.U+.!1$/+,!?Gqr!y0S4!8n+.!q!6_!.*J+.!89!^2+!.i+!`(!!
!
)'!6.c+,!>0+.!Vu!1$/+,!,_6!120!tua!
BN94)J(!rz+,!89!kJ0!67+.!.U+.!1$/+,F!8K4!

AB = a > 0
(!
rz+,!8{+.!^|!.J>!v}!6/v0+!6.*!6P6!4'>!,0P6!?GVB!qVtB!Gqt!
)N+.!8do6a!

MB = MN = a 10, BN = 2a 5
(!!!
tW+!4'>!,0P6!GVt!1$/+,!6s+!470!V!v$%!&'!

MJ ⊥ NJ
(!
qP6.!Ia!![\0!~!^J!4&$+,!80=>!8*7+!rt(!
)'!6_a!

JK //BD

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
⎧
⎨
⎪
⎪
⎩

⎪
⎪
⇒ NJ ⊥ MJ
(!!!
jde+,!4.M+,!tu!80!l$'!t!1J!1$/+,!,_6!120!Vu!6_!L.di+,!4&U+.!^J!

x −5 = 0
(!)*7!89!80=>!u!^J!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&!
x!
+,.03>!6R'!.3!

x −5 = 0
y −
7
2
= 0
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇔

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

+ (b −1)
2
= 20
(a − 5)
2
+ (b −6)
2
= 5
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇔
(b −1)
2
−(b −6)
2
= 15
(a − 5)
2
+ (b −1)
2
= 20
⎧
⎨
⎪
⎪

⎩
⎪
⎪
⇔
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

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

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)
⎧
⎨
⎪
⎪
⎩
⎪

⎪
(!
‚Y%!@AE!ƒ!@IE(0!!1J!8K4!

z = x + y.i
!4.b*!1S!4'!8do6a!

⇔ z
3
+ (1− 2i)z
2
− 2z(1+ i) − 2 = 0 ⇔ (z +1)(z
2
− 2i.z −2) = 0
⇔ (z +1)(z −1− i)(z +1−i) = 0 ⇔
z = −1
z = 1+ i
z = −1+ i
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⇔
x + yi = −1
x + yi = 1+ i
x + yi = −1+ i
⎡

⎣
⎢
⎢
⎢
⎢
⎢
⇔
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

+ 2xy.i;z
3
= x
3
− 3xy
2
+ (−y
3
+ 3x
2
y).i
(!fU!1g%!kY$!.03+!vw!k5+,!v}!
L.c6!^J!4&*+,!L.di+,!4&U+.!6_!6P6!870!^do+,!

x
2
− y
2
;x
3
− 3xy
2
; y
3
− 3x
2
y
(!!
RS%)#;@)#<T&1)#U)V[0O0!.3!L.di+,!4&U+.!

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!

P =
a
2
+ bc
a
2
+ (b + c )
2
+
b
2
+ ca
b
2
+ (c + a)
2
+
c
2
+ ab
c
2
+ (a +b)
2
−
8 3(a
2
+ b
2
+ c

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
a
2
+ (b + c )
2
(!
)di+,!4•!v$%!&'a!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&!
H!
!

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

+ (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

2
+ c
2
)
∑
(!
#$%!&'a!
!

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

2
3
(a + b + c )
2
=
18
5
.
a
2
+ b
2
+ c
2
(a + b + c )
2
(!!
!!
fU!1g%!!
!

P ≤
18
5
.
a
2
+ b
2
+ c

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

−
8 3
5
t,t = a +b + c
(!
)'!6_a

f '(t) =
72
5t
3
−
8 3
5
; f '(t) = 0 ⇔ t = 3
(!fU!‡ˆ@4E!8‰0!kY$!kdi+,!v'+,!s>! 0!80!l$'!

t = 3
(!
#$%!&'!

P ≤ f (t ) ≤ f ( 3) =
18
5
−
36
15
−
24
5

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

−
18
5
(!rY$!yŠ+,!8K4!470!

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