Đề thi thử đại học môn Toán năm 2015 (9)

Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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B4%)#%C#()DE#4$%&2:FG&!
"!
H4"I)1%J%)>K)L!ML)N=O9)P%E)Q)L4RS()/T&1)L4U&4)VE6)
DW&()L"I&X)/Y)Z[)\]^]*)
V1US)#4%)()\\^*7^.*\])
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e%f&)4g)>0&1)23)24"I)489)Q)!"#$%&'()*+,-) ).*.)Q)B4%)#%C#()hhhF6E#4$%&2:FG&))
Bi=)\)j.c*)>%d6kF)#$%!$&'!()!

y = x
3
−
x
2
2
− 2x − 2 (1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* =>'!?!7@!7AB3C!.$D3C!E!FG!$H!()!CGF!?!5&!71!IJK!71@'!F/F!.1@J!FLK!:";M!FN.!:";!.O1!0K!71@'!
P$Q3!01H.*!
Bi=).)j\c*)>%d6kF)
K; R1,1!P$AS3C!.T>3$!

sin x.(cot
2
x − cot x ) = 3(cos x − sin x )
*!
0; !R1,1!P$AS3C!.T>3$!

2
x
2
+x−1
− 2
x
2
−1
= 2
2 x
− 2
x
*!!
Bi=)7)j\c*)>%d6kF!=U3$!.UF$!P$Q3!

I =
cos2x
1+ 3(sin x −cos x )
2
dx
0
π
2
∫
*!
Bi=)l)j\c*)>%d6kF)
K; #$%!()!P$VF!W!.$%,!'X3!

z −
i.z

1+ i
=
−1+ 5i
2
*!=>'!()!P$VF!Y143!$ZP!FLK!()!P$VF!

w = (1− 2i).z
2
*!!
0; R[1!\!Y&!.]P!$ZP!F-F!()!./!3$143!C8'!^!F$_!()!?$-F!3$KJ*!#$[3!3C`J!3$143!'a.!()!.b!\M!
.U3$!c-F!(Jd.!7@!F$[3!7AZF!'a.!Ye3!$S3!<f"g*!
Bi=)])j\c*)>%d6kF!#$%!$>3$!F$GP!h*\i#j!FG!7-k!\i#j!Y&!$>3$!F$_!3$].!

AB = a,AD = 2a
*!
R[1!lMm!Yn3!YAZ.!Y&!.TJ3C!71@'!F-F!FO3$!\jMh#*!i12.!hl!5Jo3C!CGF!5e1!'p.!P$D3C!:\i#j;*!
RGF!C1_K!'p.!P$D3C!:hij;!5&!'p.!P$D3C!:\i#j;!0q3C!

60
0
*!=U3$!.$@!.UF$!?$)1!F$GP!h*\i#j!
5&!?$%,3C!F-F$!C1_K!$K1!7AB3C!.$D3C!lm!5&!hi*!!
Bi=)-)j\c*)>%d6kF!=T%3C!?$o3C!C1K3!5e1!$H!.TrF!.%O!7a!sckW!F$%!$K1!7AB3C!.$D3C!

d :
x −1
m
2
=
y − 2

−n
=
z
4
;Δ :
x −m
1
=
y
−2
=
z −1
1
!5e1!

m,n ≠ 0
*!=>'!'M3!7@!$K1!7AB3C!.$D3C!

d,Δ
(%3C!(%3C!5e1!3$KJ!?$1!7G!512.!P$AS3C!.T>3$!'p.!P$D3C!:t;!F$VK!

d,Δ
*!!!!
Bi=),)j\c*)>%d6kF!=T%3C!'p.!P$D3C!5e1!.TrF!.%O!7a!sck!F$%!$>3$!0>3$!$&3$!\i#j!FG!\:uvw";!
5&!

AB ⊥ BD
*!R[1!m!Y&!71@'!7)1!cV3C!FLK!#!IJK!j!5&!

H (

13
5
;
9
5
)
Y&!$>3$!F$12J!5Jo3C!CGF!FLK!m!
.T43!i#*!=>'!.%O!7a!F-F!7x3$!iM#Mj!012.!j!.$JaF!

x + 2y −1 = 0
*!!!
Bi=)a)j\c*)>%d6kF!R1,1!$H!P$AS3C!.T>3$!

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

⎪
⎪
⎩
⎪
⎪
⎪
⎪
(x, y ∈ !)
*!
Bi=)+)j\c*)>%d6kF!#$%!cMkMW!Y&!F-F!()!.$/F!?$o3C!Q'!.$%,!'X3!

x
2
+ y
2
+ z
2
= 2
*!=>'!C1-!.T9!3$y!
3$d.!FLK!01@J!.$VF!

P =
1
x
2
+ y
2
+
1
y

2
+ z
2
+
1
z
2
+ x
2
+
12
2− (x + y + z)
2
*!
mmm!nLmmm)
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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B4%)#%C#()DE#4$%&2:FG&!
<!
M!oV)LpB!)qrV!)estV)/uM)uV)
! !
Bi=)\)j.c*)>%d6kF)#$%!$&'!()!

y = x
3
−
x
2
2
− 2x − 2 (1)

*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* =>'!?!7@!7AB3C!.$D3C!E!FG!$H!()!CGF!?!5&!71!IJK!71@'!F/F!.1@J!FLK!:";M!FN.!:";!.O1!0K!71@'!
P$Q3!01H.*!
"* z[F!(13$!./!C1,1*!
<* =K!FG!71@'!F/F!.1@J!FLK!:";!Y&

A(1;−
7
2
)
*!
{AB3C!.$D3C!E!71!IJK!\!$H!()!CGF!?!FG!EO3C|!

y = k(x −1)−
7
2
*!
t$AS3C!.T>3$!$%&3$!7a!C1K%!71@'|!
!

x
3
−
x
2
2
− 2x − 2 = k(x −1)−
7
2

⇔ (x −1)(2x
2
+ x − 3−2k) = 0
⇔
x =1
2x
2
+ x − 3−3k = 0 (*)
⎡
⎣
⎢
⎢
*!
{@!E!FN.!:";!.O1!0K!71@'!P$Q3!01H.!?$1!:};!FG!$K1!3C$1H'!P$Q3!01H.!?$-F!"*!
!

⇔
−3k ≠ 0
Δ = 1− 8(−3− 3k) > 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇔ −
25
24
< k ≠ 0

!*!
HC#)$=;&(!~]k!

−
25
24
< k ≠ 0
*!!
Bi=).)j\c*)>%d6kF)
F; R1,1!P$AS3C!.T>3$!

sin x (cot
2
x − cot x ) = 3(cos x − sin x )
*!
E; !R1,1!P$AS3C!.T>3$!

2
x
2
+x−1
− 2
x
2
−1
= 2
2 x
− 2
x
*!!

K; {1•J!?1H3|!

sin x ≠ 0 ⇔ x ≠ kπ,k ∈ !
*!
t$AS3C!.T>3$!.AS3C!7AS3C!5e1|!
!

cot
2
x −cot x = 3(cot x −1) ⇔ cot
2
x −( 3 +1) cot x + 3 = 0
⇔
cot x =1
cot x = 3
⎡
⎣
⎢
⎢
⎢
⇔
x =
π
4
+ kπ
x =
π
6
+ kπ
⎡

⎣
⎢
⎢
⎢
⎢
⎢
⎢
*!
~]k!P$AS3C!.T>3$!FG!3C$1H'!Y&!

x =
π
4
+ kπ, x =
π
6
+ kπ,k ∈ !
*!!!
0; t$AS3C!.T>3$!.AS3C!7AS3C!5e1|!
!

(2
x
2
−1
− 2
x
)(2
x
−1) = 0 ⇔

2
x
2
−1
= 2
x
2
x
= 1
⎡
⎣
⎢
⎢
⎢
⇔
x = 0
x
2
−1 = x
⎡
⎣
⎢
⎢
⇔
x = 0
x =
1± 5
2
⎡
⎣

⎢
⎢
⎢
⎢
*!
~]k!P$AS3C!.T>3$!FG!3C$1H'!Y&!

x = 0;x =
1± 5
2
*!!
qU%)#;@)#<v&1)#w)m!R1,1!P$AS3C!.T>3$!

2
2 x
2
−x
− 2
x
2
−16.2
x
2
−x
+16 = 0
*!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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B4%)#%C#()DE#4$%&2:FG&!
v!

{€(|

x = 0;x = 1; x = −2;x = 2
*!!!!
Bi=)7)j\c*)>%d6kF!=U3$!.UF$!P$Q3!

I =
cos2x
1+ 3(sin x −cos x )
2
dx
0
π
2
∫
*!
=K!FG|!

I =
cos2x
4 −3sin 2x
dx
0
π
2
∫
*!
{p.!

t = 4 − 3sin 2x ⇒ t

2
= 4 − 3sin 2x ⇒ 2tdt = −6 cos 2xdx
*!
~>!5]k

I = −
1
3
tdt
t
−1
1
∫
= −
1
3
t
1
−1
= −
2
3
*!
Bi=)l)j\c*)>%d6kF)
K; #$%!()!P$VF!W!.$%,!'X3!

z −
i.z
1+ i
=

−1+ 5i
2
*!=>'!()!P$VF!Y143!$ZP!FLK!()!P$VF!

w = (1− 2i).z
2
*!!
0; R[1!\!Y&!.]P!$ZP!F-F!()!./!3$143!C8'!^!F$_!()!?$-F!3$KJ*!#$[3!3C`J!3$143!'a.!()!.b!\M!
.U3$!c-F!(Jd.!7@!7AZF!'a.!Ye3!$S3!<f"g*!
K; R1,!(•!

z = x + yi(x, y ∈ !)
.K!FG|!
!

x + yi −
i(x − yi)
1+ i
=
−1+ 5i
2
⇔ x + yi −
y + xi
1+ i
=
−1+ 5i
2
⇔ x + yi −
( y + xi)(1− i )
2

=
−1+ 5i
2
⇔ x + yi −
x + y + (x − y)i
2
=
−1+ 5i
2
⇔
x − y + (3y − x )i
2
=
−1+ 5i
2
⇔
x − y = −1
3y − x = 5
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇔
x =1
y = 2
⎧
⎨

⎪
⎪
⎩
⎪
⎪
⇒ z = 1+ 2i
*!
~>!5]k!

w = (1+ 2i )
2
(1− 2i ) = (1+ 2i ). (1+ 2i )(1−2i )
⎡
⎣
⎢
⎤
⎦
⎥
= 3(1+ 2i ) = 3+ 6i ⇒ w = 3−6i
*!
HC#)$=;&(!

w = 3− 6i
*!!!!
0; la.!()!.$JaF!\!FG!EO3C|!

abcd
*!
‚;!

a ∈ 1,2,3,4,5,6,7,8,9
{ }
⇒ a
FG!ƒ!F-F$!F$[3*!
‚;!

bcd
FG!

A
9
3
!F-F$!F$[3*!
~]k!.T%3C!\!FG!.d.!F,!

9.A
9
3
= 4536
()*!
R[1!„!Y&!0123!F)!F$[3!7AZF!'a.!()!.b!\!Ye3!$S3!<f"g*!
‚;!+$o3C!C1K3!'`J|!

Ω = 4536
*!
‚;!=K!.>'!()!?I!.$J]3!YZ1!F$%!„!0q3C!F-F!.>'!F-F!()!?$o3C!5AZ.!IJ-!<f"g*!

abcd ≤ 2015 ⇒ a ∈ 1,2
{ }
*!

=z"|!m2J!

a =1
M!FG!'a.!F-F$!F$[3!KM!

bcd
FG!

A
9
3
F-F$!F$[3*!
~]k!.TAB3C!$ZP!3&k!FG!

A
9
3
()*!
=z<|!m2J!

a = 2 ⇒ abcd ∈ 2013,2014,2015
{ }
FG!v!()*!
~]k!

Ω
X
= Ω −(3+ A
9
3

) = 4029
*!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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B4%)#%C#()DE#4$%&2:FG&!
^!
HC#)$=;&(!„-F!(Jd.!Fn3!.U3$!Y&

P =
4029
4536
*!!!!!!!!!
Bi=)])j\c*)>%d6kF!#$%!$>3$!F$GP!h*\i#j!FG!7-k!\i#j!Y&!$>3$!F$_!3$].!

AB = a,AD = 2a
*!
R[1!lMm!Yn3!YAZ.!Y&!.TJ3C!71@'!F-F!FO3$!\jMh#*!i12.!hl!5Jo3C!CGF!5e1!'p.!P$D3C!:\i#j;*!
RGF!C1_K!'p.!P$D3C!:hij;!5&!'p.!P$D3C!:\i#j;!0q3C!

60
0
*!=U3$!.$@!.UF$!?$)1!F$GP!h*\i#j!
5&!?$%,3C!F-F$!C1_K!$K1!7AB3C!.$D3C!lm!5&!hi*!!
!
R[1!zM+!Yn3!YAZ.!Y&!$>3$!F$12J!5Jo3C!CGF!FLK!\Ml!.T43!
ij*!
=K!FG|!

MK
AH

=
DM
DA
=
1
2
*!
=K'!C1-F!5Jo3C!\ij!FG!!

1
AH
2
=
1
AB
2
+
1
AD
2
=
1
a
2
+
1
4a
2
⇒ AH =
2a 5

5
;MK =
a 5
5
*!!
‚;!j%!ij!5Jo3C!CGF!5e1!'p.!P$D3C!:hl+;!343!

SKM
!
= 60
0
*!
hJk!TK|!

SM = MK .tan 60
0
=
a 5
5
. 3 =
a 15
5
*!
~>!5]k!

V
S .ABCD
=
1
3

SM .AB.AD =
1
3
.
a 15
5
.a.2a =
2a
3
15
15
:75 ;*!
‚;!=U3$!?$%,3C!F-F$!C1_K!lm!5&!hi*!
…]P!$H!.TrF!.%O!7a!FG!C)F!

M (0;0;0),A(0;a;0),D(0;−a;0),B (a;a;0),C (a;−a;0),S (0;0;
a 15
5
)
*!
=K!FG!m!Y&!.TJ3C!71@'!FLK!h#!343!!

N (
a
2
;−
a
2
;
a 15

10
)
*!
=K!FG|!

MN
! "!!!
= (
a
2
;−
a
2
;
a 15
10
),SB
! "!
= (a;a;−
a 15
5
),MB
! "!!
= (a;a;0)
*!
~>!5]k!

MN
! "!!!
,SB

! "!
⎡
⎣
⎢
⎤
⎦
⎥
= (0;
a
2
15
5
,a
2
);d (MN ,SB ) =
MN
! "!!!
,SB
! "!
⎡
⎣
⎢
⎤
⎦
⎥
.MB
! "!!
MN
! "!!!
,SB

! "!
⎡
⎣
⎢
⎤
⎦
⎥
=
a
3
15
5
a
4
+
15a
4
25
=
a 6
4
*!
HC#)$=;&(!!!

V
S .ABCD
=
2a
3
15

15
;d (MN ,SB ) =
a 6
4
*!!!!
Bi=)-)j\c*)>%d6kF!=T%3C!?$o3C!C1K3!5e1!$H!.TrF!.%O!7a!sckW!F$%!$K1!7AB3C!.$D3C!

d :
x −1
m
2
=
y − 2
−n
=
z
4
;Δ :
x −m
1
=
y
−2
=
z −1
1
!5e1!

m,n ≠ 0
*!=>'!'M3!7@!$K1!7AB3C!.$D3C!

d,Δ
(%3C!(%3C!5e1!3$KJ!5&!?$1!7G!512.!P$AS3C!.T>3$!'p.!P$D3C!:t;!F$VK!

d,Δ
*!!!!
{AB3C!.$D3C!E!71!IJK!71@'!l:"w<wf;!3$]3!

a
!
= (m
2
;−n;4)
Y&'!5†F!.S!F$x!P$AS3C*!
{AB3C!.$D3C!

Δ
!!71!IJK!71@'!m:'wfw";!3$]3!

b
!
= (1;−2;1)
Y&'!5†F!.S!F$x!P$AS3C*!
j%!E€€

Δ
!343!

a
!

,b
!
!F‡3C!P$AS3C!5&!

MN
! "!!!
,b
"
!?$o3C!F‡3C!P$AS3C*!
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&!
"!
#$!!!

a
!
,b
!
!%&'(!)*+,'(

⇔
m
2
1
=
−n
−2
=
4

1
⇔
m = ±2
n = 8
⎧
⎨
⎪
⎪
⎩
⎪
⎪
-!
#$!!

MN
! "!!!
,b
"
!.*/'(!%&'(!)*+,'(!

⇔
1− m
1
≠
2
−2
≠
−1
1
⇔ m ≠ 2

-!
012!*3)!*45!6578!.59'!2:;'!24!%<!=>?@A'>B-!
0*5!6<!

MN
! "!!!
= (−3;−2;1) ⇒ MN
! "!!!
,b
"
⎡
⎣
⎢
⎤
⎦
⎥
= (0;4;8) / /(0;1;2)
-!
CD2!)*E'(!FG$!%*H4!IA!

Δ
!'*J'!FKLML@$!NO=!PQ%!2,!)*R)!28S1'!PO!65!T84!C!';'!%<!)*+,'(!
2:U'*!NO!

(P ) : y + 2z − 2 = 0
-!
HC#)$=;&(!=>?@A'>B!PO!FG$V!S#@W!?@>K-!
BI=),)JKL*)>%M6NF!X:Y'(!=D2!)*E'(!PZ5!2:[%!2Y\!6]!^_S!%*Y!*U'*!`U'*!*O'*!abcd!%<!aF?eLM$!
PO!

AB ⊥ BD
-!fg5!h!NO!65i=!6j5!_H'(!%k4!c!T84!d!PO!

H (
13
5
;
9
5
)
NO!*U'*!%*518!P8/'(!(<%!%k4!h!
2:;'!bc-!XU=!2Y\!6]!%R%!6l'*!bAcAd!`512!d!2*8]%!6+m'(!2*E'(!

x + 2y −1 = 0
-!!!
)
ON)P4Q#)4%R&)#A&4)94S#)G=T&1)159(!X4!%*H'(!=5'*!an!
P8/'(!(<%!PZ5!dn-!
dY!ab!P8/'(!(<%!PZ5!bd!';'!2H!(5R%!abdh!NO!*U'*!
%*o!'*J2!FM$-!
p8S!:4!

NAB
!
= NHB
!
= 90
0
PU!PJS!2H!(5R%!abnh!']5!
251)-!q+m'(!2:r'!'OS!%*s'*!NO!6+m'(!2:r'!'(Y\5!251)!

24=!(5R%!abh!%<!2t=!NO!2:8'(!65i=!%k4!bh!F@$-!
Xu!FM$!PO!F@$!v8S!:4V!"!65i=!aAbAnAdAh!%&'(!2*8]%!
6+m'(!2:r'!6+m'(!.s'*!bh-!
wU!PJS!

AHD
!
= AND
!
= 90
0
A!2H%!an!P8/'(!(<%!PZ5!dn!
F6)%=$-!
)
ON)UV6)#"W)>X)>%M6()))
G*+,'(!2:U'*!6+m'(!2*E'(!dn!65!T84!n!PO!P8/'(!(<%!PZ5!an!%<!)*+,'(!2:U'*!
NO

7x + y − 20 = 0
-!!
XY\!6]!65i=!d!NO!'(*59=!%k4!*9!

7x + y − 20 = 0
x + 2y −1= 0
⎧
⎨
⎪
⎪
⎩
⎪

⎪
⇔
x = 3
y = −1
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ D(3;−1)
-!
#$!q+m'(!2*E'(!bc!65!T84!n!PO!vY'(!vY'(!PZ5!ad!';'!%<!)*+,'(!2:U'*!NO!

x + 3y − 8 = 0
-!
fg5!bFB?e`L`$!PZ5!`!.*R%!xy"!2*8]%!6+m'(!2*E'(!bc!24!%<!

AB
! "!!
= (11− 3b;b −1),DB
! "!!
= (5− 3b;b +1)
-!
X4!%<V!
!

AB
! "!!

.DB
! "!!
= 0 ⇔ (11− 3b)(5−3b)+ (b −1)(b +1) = 0 ⇔ b = 3(t / m);b =
9
5
(l )
-!
dY!6<!bF?MLe$-!
wU!

AB
! "!!
= DC
! "!!
⇒ C (5;1)
-!
HC#)$=;&(!wJS!bF?MLe$A!cF"LM$!PO!dFeL?M$-!!
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&!
z!
BI=)Y)JKL*)>%M6NF!f5{5!*9!)*+,'(!2:U'*!

y x + 2 − x y + 2 = 2(x
3
− y
3
)
x 2(x +
1

y
) =1+
2− y
x
3
3
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
(x, y ∈ !)
-!
q578!.59'V!

x, y ≥ −2;x, y ≠ 0;x +
1
y
≥ 0
-!
#$!h*J'!2*|S!

x = −2
*YD%!

y = −2
.*/'(!2*Y{!=}'!*9!)*+,'(!2:U'*-!
#$!~Q2!PZ5!

x > −2; y >−2
)*+,'(!2:U'*!2*H!'*|2!%k4!*9!2+,'(!6+,'(!PZ5V!

y x + 2 − x y + 2
x + 2. y + 2
=
2(x
3
− y
3
)
x + 2. y + 2
⇔
y
y + 2
−
x
x + 2
=
2(x
3
− y
3
)
x + 2. y + 2

(1)
-!
~Q2!*O=!vj!

f (t) =
t
t + 2
; f '(t) =
t + 2 −
t
2 t + 2
t + 2
=
t + 4
2 (t + 2)
3
> 0
-!
wU!PJS!'18!

x > y ⇒VT
1
(x ) = f ( y) − f (x ) < 0;VP
(1)
> 0
)*+,'(!2:U'*!P/!'(*59=-!
h18!

x < y ⇒VT
(1)

= f (y)− f (x ) > 0;VP
(1)
< 0
)*+,'(!2:U'*!P/!'(*59=-!
h*J'!2*|S!

x = y
2*Y{!=}'-!X*4S!POY!)*+,'(!2:U'*!2*H!*45!%k4!*9!24!6+3%V!!

⇔ 2x
3
+ 2x − 2 =
2
x
3
−
1
x
2
3
−1 ⇔
2x
3
+ 2x − 4
2x
3
+ 2x + 2
=
−x
3

− x + 2
x
3
(
2
x
3
−
1
x
2
3
)
2
+
1
x
.
2
x
3
−
1
x
2
3
+
1
x
2

⇔ (x
3
+ x − 2)
2
2x
3
+ 2x + 2
+
1
x
3
.
1
(
2
x
3
−
1
x
2
3
)
2
+
1
x
.
2
x

3
−
1
x
2
3
+
1
x
2
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
= 0
⇔ x
3
+ x − 2 = 0 ⇔ (x −1)(x

2
+ x + 2) = 0 ⇔ x = 1
-!
b•5!PU!

2
2x
3
+ 2x + 2
+
1
x
3
.
1
(
2
x
3
−
1
x
2
3
)
2
+
1
x
.

2
x
3
−
1
x
2
3
+
1
x
2
> 0(do x>0)
-!!
HC#)$=;&()wJS!*9!)*+,'(!2:U'*!%<!'(*59=!I8S!'*|2!

(x; y) = (1;1)
-!!
ZV&4)$=;&F!h(YO5!%R%*!%*l!:4!

x = y
!'*+!Nm5!(5{5!2:;'!24!%<!2*i!2*€%!*59'!NZ'!*3)!'*+!v48V!
!

y x + 2 − x y + 2 = 2(x
3
− y
3
)
⇔

y
2
(x + 2)− x
2
( y + 2)
y x + 2 + x y + 2
= 2(x
3
− y
3
)
⇔ (y − x)
xy + 2(x + y)
y x + 2 + x y + 2
+ 2(x
2
+ xy + y
2
)
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟

⎟
⎟
⎟
⎟
= 0
!-!
b•'(!%R%*!.12!*3)!%*H'(!=5'*!*9!%*l!%<!'(*59=!.*5!_AS!I+,'(!2u!)*+,'(!2:U'*!2*H!*45-!X8S!
'*5;'!%R%*!NO=!'OS!v‚!.*/'(!*598!T8{!'18!2u!)*+,'(!2:U'*!2*H!*45!%k4!*9!.*/'(!%*Y!)*Q)!
v8S!:4!6578!.59'!_ASƒK-!
Z[%)#;@)#<\&1)#])^)
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&!
„!
Z[%):_)*KF!f5{5!*9!)*+,'(!2:U'*!

x y + 4 − y x + 4 = 2(y − x)
x
2
+ y
2
− 2x + 2y = 8
⎧
⎨
⎪
⎪
⎩
⎪
⎪
-!qyvV!

(x; y) = (−2;−2);(2;2)
-!!!
Z[%):_)*.F)f5{5!*9!)*+,'(!2:U'*!

y x + 2 − x y + 2 = 2(x
3
− y
3
)
x
3
− 4x
2
+ 4x −1 = 13− y
2
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
-!qyvV!

(x; y) = (3;3)
-!
ndV!XU=!6+3%!

y = x
6+4!P7!(5{5!)*+,'(!2:U'*V!!
!

x
3
− 4x
2
+ 4x −1 = 13− x
2
⇔ (x
3
− 4x
2
+ 4x − 3) + (2− 13− x
2
) = 0
⇔ (x − 3)(x
2
− x +1)+
x
2
− 9
13− x
2
+ 2
= 0
⇔ (x − 3)(x
2

− x +1+
x + 3
13− x
2
+ 2
) = 0 ⇔ x = 3
!-!
IY!

x
2
− x +1+
x + 3
13− x
2
+ 2
> 0,∀x >−2
-!!
HC#)$=;&(!n9!%<!'(*59=!I8S!'*|2!

(x; y) = (3;3)
-!!
BI=)+)JKL*)>%M6NF!c*Y!_ASAW!NO!%R%!vj!2*€%!.*/'(!t=!2*Y{!=}'!

x
2
+ y
2
+ z
2

= 2
-!XU=!(5R!2:…!'*†!
'*|2!%k4!`5i8!2*H%!

P =
1
x
2
+ y
2
+
1
y
2
+ z
2
+
1
z
2
+ x
2
+
12
2− (x + y + z)
2
-!
dY!

x

2
+ y
2
+ z
2
= 2 ⇒ 2−(x + y + z)
2
= −2(xy + yz + zx )
-!
p8S!:4V

P =
1
x
2
+ y
2
+
1
y
2
+ z
2
+
1
z
2
+ x
2
−

6
xy + yz + zx
-!
X4!%<V!
!

1
x
2
+ y
2
∑
=
1
2
x
2
+ y
2
+ z
2
x
2
+ y
2
∑
=
3
2
+

1
2
z
2
x
2
+ y
2
∑
-!
X4!%<V!
!

(
z
2
x
2
+ y
2
∑
)(x
2
y
2
+ y
2
z
2
+ z

2
x
2
) ≥ x
4
+ y
4
+ z
4
-!
wU!PJS!!

1
x
2
+ y
2
∑
≥
3
2
+
1
2
.
x
4
+ y
4
+ z

4
x
2
y
2
+ y
2
z
2
+ z
2
x
2
=
1
2
+
(x
2
+ y
2
+ z
2
)
2
2(x
2
y
2
+ y

2
z
2
+ z
2
x
2
)
≥
1
2
+
(x
2
+ y
2
+ z
2
)
2
2(xy + yz + zx )
2
=
1
2
+
2
(xy + yz + zx )
2
≥

6
xy + yz + zx
−4
-!
d|8!`•'(!6\2!%*E'(!*\'!

x = 0;x
2
+ y
2
+ z
2
= 2;xy + yz + zx =
1
3
-!
wJS!(5R!2:…!'*†!'*|2!%k4!G!`•'(!?‡-!!!!
b•'(!%R%*!2+,'(!2€!24!%<!.12!T8{!2ˆ'(!T8R2!v48V!
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&!
B!

1
a
2
+ b
2
+
1

b
2
+ c
2
+
1
c
2
+ a
2
+
k
a
2
+ b
2
+ c
2
≥
α
k
ab + bc + ca
-!
X:Y'(!6<!
!

α
k
= (2k + 9) / 2 for k ≤−4,
α

k
= (k + 5) / 2 for −4 ≤ k ≤ 3,
α
k
= 2 k +1 for k ≥ 3
!!
Z[%)#;@)#<\&1)#])
Z[%):_)*KF!c*Y!4A`A%!NO!%R%!vj!2*€%!.*/'(!t=!2*Y{!=}'!

ab + bc + ca =1
-!XU=!(5R!2:…!'*†!'*|2!
%k4!`5i8!2*H%!

P =
1
a
2
+ b
2
+
1
b
2
+ c
2
+
1
c
2
+ a

2
+
1
a
2
+ b
2
+ c
2
-!
Z[%):_)*.F!c*Y!4A`A%!NO!%R%!vj!2*€%!.*/'(!t=!2*Y{!=}'!

a
2
+ b
2
+ c
2
= 2
-!XU=!(5R!2:…!'*†!'*|2!
%k4!`5i8!2*H%!

P =
1
a
2
+ b
2
+
1

b
2
+ c
2
+
1
c
2
+ a
2
−
4
ab + bc + ca
-!
BI=)+)JKL*)>%M6NF!c*Y!4A`A%!NO!%R%!vj!2*€%!.*/'(!t=!%*H'(!=5'*!:•'(!
!

1
a
2
+ b
2
+
1
b
2
+ c
2
+
1

c
2
+ a
2
+
8
a
2
+ b
2
+ c
2
≥
6
ab + bc + ca
-!
!
!
!!
!!
!!!!
!!
!!
!
!!!!!!!