đề thi thử thpt quốc gia môn toán,đề số 17

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"!
Khoá%giải%đề%THPT%Quốc%Gia%–%Thầy:%Đặng%Thành%Nam%
Môn:%Toán;%ĐỀ%SỐ%17/50%
Ngày%thi%:%18/03/2015%
Thời%gian%làm%bài:%180%phút,%không%kể%thời%gian%giao%đề%
Liên%hệ%đăng%ký%khoá%học%–%Hotline:%0976%266%202%–%Chi%tiết:%www.mathlinks.vn%%
Câu%1%(2,0%điểm).%#$%!$&'!()!

y =
x + 2− m
x + m
2
(1)
!*+,!'!-&!.$/'!()!.$012!

m ≠1;m ≠ −2
3!
"3 4$5%!(6.!(0!7,89!.$,:9!*&!*;!<=!.$>!$&'!()!?"@!*+,!

m = −1
3!
A3 BC'!'!<D!$&'!()!?"@!9E$>1$!7,89!.F:9!G$%59E!

(−4;+∞)
3!
Câu%2%(1,0%điểm).%
/@ H,5,!I$JK9E!.FC9$!

log
25
(x +1)+
1
2
log
5
6x +1 =1
3!
7@ BC'!E,6!.F>!-+9!9$L.!*&!9$M!9$L.!1N/!$&'!()!

y = log
2
(x +1)− x
3
!.F:9!<%O9!PQR"S3!!
Câu%3%(1,0%điểm).!BT9$!.T1$!I$U9!

I =
x
3
−2x + (x −1)e
x
e
x
+ x
2
dx
0
1

∫
3!
Câu%4%(1,0%điểm).%
/@ #$%!()!I$V1!

z = cos x − i.sin x
3!BC'!()!.$01!W!(/%!1$%!

z −i = 2
3!
7@ BC'!()!$O9E!1$V/!

x
16
!.F%9E!G$/,!.F,D9!

(x
2
−
1
x
+1)
n
!7,8.!9!-&!()!.0!9$,:9!.$%5!'X9!

C
n
2
= n + 27
3!

Câu%5%(1,0%điểm).!#$%!$C9$!1$YI!Z3[\#]!1Y!<6^![\#]!-&!$C9$!7C9$!$&9$!1Y!

AB = a,AD = a 3, BAD
!
= 30
0
3!_C9$!1$,8`!*`a9E!EY1!1N/!Z!-:9!'b.!I$c9E!?[\#]@!.Fd9E!*+,!
.F`9E!<,D'!1O9$![]2!EY1!E,e/!Z#!*&!'b.!I$c9E!?[\#]@!7f9E!

60
0
3!BT9$!.$D!.T1$!G$),!1$YI!
Z3[\#]!*&!G$%59E!161$!.g!<,D'![!<89!'b.!I$c9E!?Z\]@3!
Câu%6%(1,0%điểm).!BF%9E!G$a9E!E,/9!*+,!$h!.Fi1!.%O!<j!kW^l!1$%!'b.!I$c9E!

(P ) : 2x + y + 2z −14 = 0
3!B/'!E,61![\#!1U9!.O,![!1Y!\2#!.$`j1!'b.!I$c9E!?m@!*&!9$n9!H?oRpR"@!
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.$c9E!\#3!
Câu%7%(1,0%điểm).!BF%9E!'b.!I$c9E!*+,!.Fi1!.%O!<j!kW^!1$%!./'!E,61!9$q9![\#!

(AC > AB )
3!
Hq,!

D(2;−
3
2
)
!-&!1$U9!<Jw9E!I$U9!E,61!.F%9E!EY1![2!x?u"RQ@!-&!'j.!<,D'!.$`j1!<%O9![#!.$%5!

'X9!

AB = AE
3!BC'!.%O!<j!161!<y9$![2\2#!7,8.!I$JK9E!.FC9$!<Jw9E!.Fz9!9E%O,!.,8I!./'!E,61!
[\#!-&!

x
2
+ y
2
+ x − 2y −30 = 0
!*&![!1Y!$%&9$!<j!{JK9E3!
Câu%8%(1,0%điểm).%H,5,!$h!I$JK9E!.FC9$

x + y
2
+
x
2
+ y
2
2
= 2
x
3
+ y
3
2
3
8x

3
− 4 x + 3 + 6 = 3
x + y
2
( x + y )
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
(x, y ∈ !)
3!
Câu%9%(1,0%điểm).%#$%!/2721!-&!161!()!.$01!G$a9E!U'!.$%5!'X9!

a
2
+ b
2
+ c
2

= 2
!*&!

a ≥ b
3!BC'!E,6!
.F>!-+9!9$L.!1N/!7,D`!.$V1!

P =
1
a
3
+ b
2
+ c
+
1
b
3
+ c
2
+ a
−
5(a + c )(a + c −1)
4
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A!
PHÂN%TÍCH%BÌNH%LUẬN%ĐÁP%ÁN%
Câu%1%(2,0%điểm).%#$%!$&'!()!

y =
x + 2− m
x + m
2
(1)
!*+,!'!-&!.$/'!()!.$012!

m ≠1;m ≠ −2
3!
"3 4$5%!(6.!(0!7,89!.$,:9!*&!*;!<=!.$>!$&'!()!?"@!*+,!

m = −1
3!
A3 BC'!'!<D!$&'!()!?"@!9E$>1$!7,89!.F:9!G$%59E!

(−4;+∞)
3!
"3 _q1!(,9$!.0!E,5,3!
A3 BnI!W61!<>9$|!

D = !\ −m
2
{ }
3!B/!1Y|!

y ' =
m
2
+ m − 2

(x + m
2
)
2
3!
}!@!_&'!()!9E$>1$!7,89!.F:9!G$%59E!

(−4;+∞)
G$,!*&!1$y!G$,|!

x + m
2
≠ 0
y ' <0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
,∀x ∈ (−4;+∞) ⇔
m
2
+ m − 2 < 0
−m
2
∈ −4;+∞
⎡
⎣

⎢
)
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇔ −2 <m <1
3!
Kết%luận:%vn^!

−2 < m <1
-&!E,6!.F>!1~9!.C'3!!
Câu%2%(1,0%điểm).%
/@ H,5,!I$JK9E!.FC9$!

log
25
(x +1)+
1
2
log
5
6x +1 =1
3!
7@ BC'!E,6!.F>!-+9!9$L.!*&!9$M!9$L.!1N/!$&'!()!

y = log
2
(x +1)− x
3
!.F:9!<%O9!PQR"S3!!
/@ •,€`!G,h9|!

x > −
1
6
3!
m$JK9E!.FC9$!.JK9E!<JK9E!*+,|!

1
2
log
5
(x +1)+
1
2
log
5
6x +1 =1 ⇔ log
5
(x +1)+ log
5
6x +1 = 2
⇔ log
5

(x +1) 6x +1 = 2 ⇔ (x +1) 6x +1 = 25
⇔ (x +1)( 6x +1 −5)+5(x −4) = 0
⇔
6(x +1)(x − 4)
6x +1 + 5
+ 5(x −4) = 0 ⇔ (x − 4)
6(x +1)
6x +1 + 5
+ 5
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
= 0 ⇔ x = 4
3!
Kết%luận:!m$JK9E!.FC9$!1Y!9E$,h'!{`^!9$L.!

x = 4
3!!
#$•!‚!1Y!.$D!.$01!$,h9!161!161$!G$61!(/`!<U^|!
}@!_&'!()!

y = log
25

(x +1)+
1
2
log
5
6x +1
<=9E!7,89!9:9!I$JK9E!.FC9$!1Y!.),!</!'j.!9E$,h'2!
'b.!G$61!9$n9!.$L^!Wƒs!.$%5!'X9!I$JK9E!.FC9$3!vn^!9:9!I$JK9E!.FC9$!1Y!9E$,h'!{`^!9$L.!
Wƒs3!
}@!H,5,!I$JK9E!.FC9$|!!

(x +1) 6x +1 = 25 ⇔ (x +1)
2
(6x +1) = 25
2
⇔ (x − 4)(6x
2
+ 37x +156) = 0 ⇔ x = 4
3!
7@ !_&'!()!<X!1$%!-,:9!.i1!.F:9!<%O9!PQR"S3!
B/!1Y|!

y ' =
1
(x +1).ln 2
−3x
2
≤
1
2ln 2

−3< 0
3!
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o!
vn^!$&'!()!9E$>1$!7,89!.F:9!<%O9!PQR"S3!
]%!<Y!

y
min
= y(1) = 0; y
max
= y(0) = 0
3!!!
Câu%3%(1,0%điểm).!BT9$!.T1$!I$U9!

I =
x
3
−2x + (x −1)e
x
e
x
+ x
2
dx
0
1
∫

3!

I =
x(x
2
+ e
x
)−(2x + e
x
)
e
x
+ x
2
dx
0
1
∫
= (x −
2x + e
x
e
x
+ x
2
)dx
0
1
∫
= (

x
2
2
−ln e
x
+ x
2
)
1
0
=
1
2
−ln(e +1)
3!
Kết%luận:!vn^!

I =
1
2
− ln(e +1)
3!!
Câu%4%(1,0%điểm).%
/@ #$%!()!I$V1!

z = cos x − i.sin x
3!BC'!()!.$01!W!(/%!1$%!

z −i = 2
3!

7@ BC'!()!$O9E!1$V/!

x
16
!.F%9E!G$/,!.F,D9!

(x
2
−
1
x
+1)
n
!7,8.!9!-&!()!.0!9$,:9!.$%5!'X9!

C
n
2
= n + 27
3!
/@ B/!1Y|!

z −i = cos x + (sin x +1).i ⇒ z − i = cos
2
x + (sin x +1)
2
3!!
}@!B$„%!E,5!.$,8.!./!1Y|!
!

cos
2
x + (sin x +1)
2
= 2 ⇔ cos
2
x + (sin x +1)
2
= 4
⇔ 1−sin
2
x + (sin x +1)
2
= 4 ⇔ 2sin x + 2 = 4 ⇔ sin x = 1⇔ x =
π
2
+ k2π
3!
48.!-`n9|!vn^!

x =
π
2
+ k2π,k ∈ !
3!!!
7@ B$„%!E,5!.$,8.!./!1Y!I$JK9E!.FC9$|!
!

n!
2!(n −2)!

= n + 27 ⇔
n(n −1)
2
= n + 27
⇔ n
2
−3n −54 = 0 ⇔
n = 9(t / m)
n = −6(l )
⎡
⎣
⎢
⎢
3!
vn^!

n = 9
*&!

(x
2
−
1
x
+1)
9
= C
9
k
(x

2
−
1
x
)
k
k=0
9
∑
= C
9
k
. (−1)
i
x
2(k−i )
.x
−i
i=0
k
∑
k=0
9
∑
= (−1)
i
.x
2k−3i
i=0
k

∑
k=0
9
∑
3!
B/!1~9!1$q9!G2,!.$%5!'X9!

0 ≤ i ≤ k ≤ 9
2k −3i = 16
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ (i;k ) = (0;8)
3!
vn^!()!$O9E!1~9!.C'!-&!

C
9
8
.(−1)
0
x
16
= 9x
16
3!!

Câu%5%(1,0%điểm).!#$%!$C9$!1$YI!Z3[\#]!1Y!<6^![\#]!-&!$C9$!7C9$!$&9$!1Y!

AB = a,AD = a 3, BAD
!
= 30
0
3!_C9$!1$,8`!*`a9E!EY1!1N/!Z!-:9!'b.!I$c9E!?[\#]@!.Fd9E!*+,!
KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%%
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s!
.F`9E!<,D'!1O9$![]2!EY1!E,e/!Z#!*&!'b.!I$c9E!?[\#]@!7f9E!

60
0
3!BT9$!.$D!.T1$!G$),!1$YI!
Z3[\#]!*&!G$%59E!161$!.g!<,D'![!<89!'b.!I$c9E!?Z\]@3!
!
Hq,!_!-&!.F`9E!<,D'!1O9$![]2!./!1Y|!

SH ⊥ (ABCD )
3!
v&!_#!-&!$C9$!1$,8`!*`a9E!EY1!1N/!Z#!.F:9!'b.!I$c9E!
?[\#]@!9:9!

SCH
!
= 60
0
3!!

]%!

BAD
!
= 30
0
⇒ ADC
!
= 150
0
3!
…I!{i9E!<>9$!-‚!$&'!()!#a(,9!1$%!./'!E,61!_]#!1Y|!
!

CH = HD
2
+CD
2
− 2HD.CD cos150
0
=
a
2
4
+ 3a
2
− 2.
a
2
.a 3.(−

3
2
) =
a 19
2
3!!
Z`^!F/|!

SH = HC.tan 60
0
=
a 19
2
. 3 =
a 57
2
3!!
}@!

S
ABCD
= 2S
ABD
= AB.AD.sin30
0
= a.a 3.
1
2
=
a

2
3
2
3!
vC!*n^!

V
S .ABCD
=
1
3
SH .S
ABCD
=
1
3
.
a 57
2
.
a
2
3
2
=
a
3
19
4
?<* @3!

}@!BT9$!G$%59E!161$!.g![!<89!'b.!I$c9E!?Z\]@3!
B/!1Y!

d (A;(SBD)) =
AD
HD
.d(H ;(SBD )) = 2.d(H ;(SBD))
3!
4†!_‡!*`a9E!EY1!*+,!\]!.O,!‡!*&!G†!_4!*`a9E!EY1!*+,!Z‡!.O,!4!./!1Y!

HK ⊥ (SBD )
3!
}@!…I!{i9E!<>9$!-‚!$&'!()!1a!(,9!1$%!./'!E,61![\]!1Y!!
!

BD = AB
2
+ AD
2
− 2AB.AD cos30
0
= a
2
+ 3a
2
− 2a.a 3.
3
2
= a
!3!

B/!1Y!

HI =
2S
HBD
BD
=
S
ABD
BD
=
S
ABCD
2BD
=
a
2
3
4a
=
a 3
4
3!
B/'!E,61!*`a9E!Z_‡!1Y|!
!

1
HK
2
=

1
SH
2
+
1
HI
2
=
16
3a
2
+
4
57a
2
=
308
57a
2
⇒ HK =
a
2
57
77
3!
Z`^!F/|!

d (A;(SBD)) = a
57
77

3!!!!!!
Câu%6%(1,0%điểm).!BF%9E!G$a9E!E,/9!*+,!$h!.Fi1!.%O!<j!kW^l!1$%!'b.!I$c9E!

(P ) : 2x + y + 2z −14 = 0
3!B/'!E,61![\#!1U9!.O,![!1Y!\2#!.$`j1!'b.!I$c9E!?m@!*&!9$n9!H?oRpR"@!
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.$c9E!\#3!
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Chi%tiết:%Mathlinks.vn!
"!
#$!%&!'!()!*+, !*/0!*10!.234!567!-8-!

AG
! "!!
= 2GM
! "!!
= (2;4;−4) ⇒
3− x
A
= 2
6− y
A
= 4
1− z
A
= −4
⎧
⎨
⎪

⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⇒ A(1;2;5)
9!
#$!:;!*10!.234!567!4/-!*<2!5!-8-!

BC ⊥ AM
9!
=1!4>?!

AM
! "!!!
= (3;6;−6) //(1;2;−2),n
P
!"!
= (2;1;2) ⇒ AM
! "!!!
,n
P
!"!
⎡
⎣
⎢
⎤

⎦
⎥
= (6;−6;−3) //(2;−2;−1)
9!
=1!4>!

BC ⊥ AM ,BC ⊥ n
P
!"!
!@AB!+1!67!-CD-!EFGHFGHI$!()0!JK4!*L!4CM!NCOL 9!
:;!P>!67!P2!QA1!R!4>!NCOL !*+&-C!()!

x − 4
2
=
y −8
−2
=
z +1
−1
9!!!!
Kết%luận:!5EIGFG"$!J)!

BC :
x − 4
2
=
y −8
−2
=

z +1
−1
9!
Câu%7%(1,0%điểm).!=+; !0S*!NCT !JU2!*+V4!*;<!PW!XYB!4C;!*10!.234!-C,-!567!

(AC > AB )
9!
',2!

D(2;−
3
2
)
!()!4C/-!POZ !NC/-!.234!*+; !.>4!5[!\EHIG]$!()!0W*!P2^0!*CAW4!P;<-!57!*C;_!
0`-!

AB = AE
9!=&0!*;<!PW!434!PM-C!5[6[7!a2b*!NCOL !*+&-C!POZ !*+c-! ;<2!*2bN!*10!.234!
567!()!

x
2
+ y
2
+ x − 2y −30 = 0
!J)!5!4>!C;)-C!PW!dOL 9!
!
eOZ !*+c-! ;<2!*2bN!*10!.234!567!4>!*/0!

I (−

1
2
;1)
![!
a3-!fg-C!ah !

5 5
2
9!
',2!i!()!.21;!P2^0!4j1!5k!J)!POZ !*CT !:\9!
#$!lK*!C12!*10!.234!56:!J)!5\:!4>?!
#!

AB = AE,BAD
!
= EAD
!
[!5:!4CA !-8-!*10!.234!56:!
ah !*10!.234!5:\9!!
mAB!+1?!

AED
!
= ABC
!
9!
=1!4>?!

HAE
!

= ICA
!
=
180
0
− AIC
!
2
= 90
0
− ABC
!
9!
mAB!+1?!

AHE
!
= AED
!
+ HAE
!
= ABC
!
+ (90
0
− ABC
!
) = 90
0
9!%&!JDB!5k!JAn !.>4!JU2!POZ !*CT !

:\9!!
#$!=1!4>?!

DE
! "!!
= (−3;
3
2
) //(2;−1)
9!!!!
#$!eOZ !*CT !5k!P2!QA1!k!J)!JAn !.>4!JU2!:\!-8-!4>!NCOL !*+&-C!()!

2x − y + 2 = 0
9!
=;<!PW!P2^0!5!()! C2o0!4j1!Co!

2x − y + 2 = 0
x
2
+ y
2
+ x − 2y −30 = 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇔

x = 2, y = 6
x = −3,y = −4
⎡
⎣
⎢
⎢
⇒ A(2;6)
9!
#$!eOZ !*CT !5:!P2!QA1!5[:!4>!NCOL !*+&-C!()!

x − 2 = 0
9!
',2!5p!()!.21;!P2^0!*Cq!C12!4j1!5:!J)!POZ !*+c-!E7$[!*;<!PW!P2^0!5p!*C;_!0`-!
KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%%
Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%%
Chi%tiết:%Mathlinks.vn!
r!
!

x = 2
x
2
+ y
2
+ x − 2y −30 = 0
⎧
⎨
⎪
⎪
⎩

⎪
⎪
⇔
x = 2, y = 6
x = 2, y = −4
⎡
⎣
⎢
⎢
⇒ A'(2;−4)
9!
#$!eOZ !*CT !67!P2!QA1!:!J)!JAn !.>4!JU2!k5p!4>!NCOL !*+&-C!()!
!

x − 2 y −5 = 0
9!
=;<!PW!P2^0!6[7!*C;_!0`-!Co!

x −2y −5 = 0
x
2
+ y
2
+ x − 2y −30 = 0
⎧
⎨
⎪
⎪
⎩
⎪

⎪
⇔
x = 5, y = 0
x = −3, y = −4
⎡
⎣
⎢
⎢
9!
mAB!+1!6E"G]$[!7EHsGHt$!C;S4!6EHsGHt$[!7E"G]$9!
eu2!4C2bA!JU2!P2vA!f2o-!57w56!*1!4>?!6E"G]$!J)!7EHsGHt$9!
Kết%luận:!%DB!

A(2;6),B(5;0),C (−3;−4)
9!!
!
Câu%8%(1,0%điểm).%'2_2!Co!NCOL !*+&-C

x + y
2
+
x
2
+ y
2
2
= 2
x
3
+ y

3
2
3
8x
3
− 4 x + 3 + 6 = 3
x + y
2
( x + y )
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
(x, y ∈ !)
9!
e2vA!f2o-?!

x, y ≥ 0
9!

=1!4>?!
!

x
2
+ y
2
2
≥
1
4
(x + y)
2
=
x + y
2
9!
:;!P>!

VT
(1)
≤ 2
x
2
+ y
2
2
⇒VP
(1)
= 2

x
3
+ y
3
2
3
≤ 2
x
2
+ y
2
2
9!
mAB!+1?!
!

(
x
3
+ y
3
2
)
2
≤ (
x
2
+ y
2
2

)
3
⇔ 2(x
3
+ y
3
)
2
−(x
2
+ y
2
)
3
≤ 0
⇔ (x − y)
2
(x
4
+ 2x
3
y + 2xy
3
+ y
4
) ≤ 0 ⇔ x = y
9!
:;!JDB!NCOL !*+&-C!PxA!4j1!Co!*OL !POL !JU2?!

y = x

9!
=C1B!J);!NCOL !*+&-C!*Cq!C12!4j1!Co!*1!POy4?!
!

4x
3
− 2 x + 3 + 3= 3x ⇔ 4x
3
− 3x −1+ 2(2− x + 3) = 0
⇔ (x −1)(4x
2
+ 4x +1)+
2(1− x)
2+ x + 3
= 0
⇔ (x −1)(4x
2
+ 4x +1−
2
2+ x + 3
) = 0
⇔ (x −1)(8x
2
+ 8x + (2x +1)
2
x + 3) = 0 ⇔ x = 1
9!
6z2!J&!

8x

2
+ 8x + (2x +1)
2
x + 3 > 0,∀x ≥ 0
9!
Kết%luận:!io!NCOL !*+&-C!4>! C2o0!dAB!-C{*!

(x; y) = (1;1)
9!!!!!!
Cách%2:!|CD-!*C{B!NCOL !*+&-C!PxA!4j1!Co!4>!d< !PT !4{N!-8-!*1!4>!*C^!PS*!B}*9Y!PO1!Jv!
.2_2!NCOL !*+&-C!JU2!*9!
KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%%
Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%%
Chi%tiết:%Mathlinks.vn!
~!
|CD-!*C{B!

x = 0
fCn !()! C2o0!4j1!Co!-8-!YK*!JU2!

x > 0
PS*!

t =
y
x
≥ 0
NCOL !*+&-C!*Cq!-C{*!
4j1!Co!*+z!*C)-C?!!
=1!4>?!

!

(t
3
+1)(t
3
+1)(1+1) ≥ (t
2
+1)
3
⇒
t
3
+1
2
3
≥
t
2
+1
2
≥
t +1
2
⇒VT ≤VP
9!
:{A!ah !Y_B!+1!fC2!J)!4CM!fC2!

t = 1
9!!

%&!JDB!NCOL !*+&-C!PxA!4j1!Co!*1!4>?!

y = x
9!
Bình%luận:!=1!4>!*C^!@AB!+1!a{*!PT !*Cq4!*+8-!-CO!@1A?!
!

2(t
3
+1)
2
−(t
2
+1)
3
= (t −1)
2
(t
4
+ 2t
3
+ 2t +1) ≥ 0,∀t ≥ 0
9!
:;!P>!

t
3
+1
2
3

≥
t
2
+1
2
9!!!!
#$!=• !QA3*!*1!4>?!
!

(t
n
+1)(t
n
+1)(1+1) (1+1)
(n−2) lân
! "##### $#####
≥ (t
2
+1)
n
⇒
t
n
+1
2
n
≥
t
2
+1

2
9!
:;!P>!NCOL !*+&-C!PxA!4j1!Co!4>!*C^!4C;!d< !*• !QA3*?!
!

x + y
2
+
x
2
+ y
2
2
= 2.
x
n
+ y
n
2
n
9!
Bài%tập%tương%tự%}%%
Bài%số%01.%'2_2!Co!NCOL !*+&-C!

x + y
2
+
x
2
+ y

2
2
= 2
x
4
+ y
4
2
4
x
3
−3x −1 = 8 − 3y
2
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
9!!
Câu%9%(1,0%điểm).%7C;!1[a[4!()!434!@u!*C€4!fCn !/0!*C;_!0`-!

a
2
+ b

2
+ c
2
= 2
!J)!

a ≥ b
9!=&0!.23!
*+•!(U-!-C{*!4j1!a2^A!*Cq4!

P =
1
a
3
+ b
2
+ c
+
1
b
3
+ c
2
+ a
−
5(a + c )(a + c −1)
4
9! !
m‚!dV !a{*!PT !*Cq4!71A4CB!ƒm4C„1+…!*1!4>?!

(a
3
+ b
2
+ c )(a + b
2
+ c
3
) ≥ (a
2
+ b
2
+ c
2
)
2
,
⇒
1
a
3
+ b
2
+ c
≤
a + b
2
+ c
3
(a

2
+ b
2
+ c
2
)
2
;
(b
3
+ c
2
+ a)(b + c
2
+ a
3
) ≥ (a
2
+ b
2
+ c
2
)
2
,
⇒
1
b + c
2
+ a

3
≤
b + c
2
+ a
3
(a
2
+ b
2
+ c
2
)
2
9!
%&!JDB!
KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%–%Thầy:%ĐẶNG%THÀNH%NAM%%
Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%%
Chi%tiết:%Mathlinks.vn!
†!
!

1
a
3
+ b
2
+ c
+
1

b
3
+ c
2
+ a
≤
a + b
2
+ c
3
(a
2
+ b
2
+ c
2
)
2
+
b + c
2
+ a
3
(a
2
+ b
2
+ c
2
)

2
=
a + b + b
2
+ c
2
+ a
3
+ c
3
4
≤
a + b + (b + c )
2
+ (a + c )
3
4
≤
a + b + (a + c)
2
+ (a + c )
3
4
9!
mAB!+1?!
!

P ≤
a + b + (a + c)
2

+ (a + c )
3
4
−
5(a + c )(a + c −1)
4
=
(a + c)
3
− 4(a + c )
2
+ 5(a + c ) + a +b
4
9!
=1!4>?!

a + b ≤ 2(a
2
+ b
2
) ≤ 2(a
2
+ b
2
+ c
2
) = 2
9!
%)!PS*!

x = a + c ⇒ f (x ) = x
3
− 4x
2
+ 5x ≤ 2
9!
%&!JDB!

P ≤
a + b + f (x)
4
≤
2+ 2
4
= 1
9!:{A!ah !P<*!*<2!

a = b = 1,c = 0
9!
%DB!.23!*+•!(U-!-C{*!4j1!‡!ah !I9!!!!!
6h !434C!*OL !*€!*1!4Cq !02-C!POy4?!
Bài%số%01.%7C;!1[a[4!()!434!@u!*C€4!dOL !*C;_!0`-!

a + b + c = 3
9!7Cq !02-C!+h !
!

1
a
3

+ b
2
+ c
+
1
b
3
+ c
2
+ a
+
1
c
3
+ a
2
+ b
≤1
9!
!
!!!!
!
!
!!