đề thi thử thpt quốc gia môn toán,đề số 14

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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)
Bj=)\)k.d*)>%e6lF!#$%!$&'!()!

y =
3x +1
x −1
(1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* =12.!>$?@3A!.BC3$!.12>!.DE23!FGH!:";!012.!.12>!.DE23!FI!$J!()!AIF!0K3A!

−1
*!!
Bj=).)k\d*)>%e6lF))
H; L1,1!>$?@3A!.BC3$!

cos3x − 4 cos2x + 8 cos x − 5 = 0
*!!
0; L1,1!0M.!>$?@3A!.BC3$!

1
log

x
2
+ 2log
2x
2 ≥ 2
*!!
Bj=)7)k\d*)>%e6lF!NO3$!.OF$!>$P3!

I =
e
x
e
x
+ 2 e
x
+ 3
dx
0
ln 6
∫
*!
Bj=)])k\d*)>%e6lF)))
H; LQ1!

z
1
,z
2
R&!$H1!3A$1J'!FGH!>$?@3A!.BC3$!

z
2
− 3z + 5 = 0
*!NC'!>$S3!.$/F!5&!>$S3!,%!FGH!()!
>$TF!

w = (z
1
2
+ z
2
2
).i + 2−3i
*!!
0; LQ1!U!R&!.V>!$W>!F-F!()!./!3$143!FI!X!F$Y!()!5&!F$1H!$2.!F$%!Z*!NO3$!()!>$S3!.[!FGH!U!5&!
.\3A!FGH!.M.!F,!F-F!>$S3!.[!7I*!
Bj=)_)k\d*)>%e6lF!#$%!$C3$!F$I>!.H'!A1-F!7]D!^*_`#!FI!7-E!_`#!R&!.H'!A1-F!7]D!Fa3$!<H*!LQ1!
bcd!RS3!R?W.!R&!.BD3A!71e'!FGH!F-F!7%a3!`#c^b*!`12.!_d!5Df3A!AIF!5g1!'h.!>$i3A!:^`#;*!
NO3$!.$e!.OF$!j$)1!F$I>!^*_`#!5&!j$%,3A!F-F$!A1YH!$H1!7?k3A!.$i3A!^_!5&!`d*!
Bj=)-)k\d*)>%e6lF!NB%3A!j$f3A!A1H3!5g1!$J!.BlF!.%a!7m!noEp!F$%!$C3$!0C3$!$&3$!_`#q!FI!
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F$TH!7?k3A!.$i3A!n`!5g1!n!R&!A)F!.%a!7m*!
Bj=),)k\d*)>%e6lF!NB%3A!'h.!>$i3A!5g1!.BlF!.%a!7m!noE!F$%!$C3$!.$%1!_`#q!FI!w1J3!.OF$!0K3A!

2 3
c!>$?@3A!.BC3$!7?k3A!F$x%!`q!R&!

3x − y = 0
*!yz3$!_!R&!$C3$!F$12D!5Df3A!AIF!FGH!`!.B43!
7?k3A!.$i3A!

d : 3x + y = 0
*!NC'!.%a!7m!0)3!7z3$!_c`c#cq!012.!7z3$!_!FI!$%&3$!7m!w?@3A*!
Bj=)`)k\d*)>%e6lF!L1,1!$J!>$?@3A!.BC3$!

(x −1)
3
− 2x( y + 3) + 2y − 2 = 0
(x + y)
2
+ y(1− −2x −1) = x
2
− (−2x −1)
3
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
(x, y ∈ ! )
*!
Bj=)+)k\d*)>%e6lF)#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!.$%,!'{3!

a
2
(

a
c
+1)+ b
2
(
b
c
+1) = 3
*!NC'!A1-!.B9!
3$|!3$M.!FGH!01eD!.$TF

P =
a + c
b
3
+ 2
+
b + c
a
3
+ 2
− 2 a + b + c
*!
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&!
<!

)
PHÂN TÍCH BÌNH LUẬN ĐÁP ÁN
Bj=)\)k.d*)>%e6lF!#$%!$&'!()!

y =
3x +1
x −1
(1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* =12.!>$?@3A!.BC3$!.12>!.DE23!FGH!:";!012.!.12>!.DE23!FI!$J!()!AIF!0K3A!

−1
*!!
"* }QF!(13$!./!A1,1*!
<* L1,!([!.12>!71e'!

M (m;
3m +1
m −1
),m ≠1
*!!
}J!()!AIF!FGH!.12>!.DE23!.a1!b!R&!

k = y'(m) = −
4
(m −1)
2
*!
N$~%!A1,!.$12.!.H!FI!>$?@3A!.BC3$•!!!

−
4
(m −1)
2
= −1 ⇔ (m −1)
2
= 4 ⇔
m = −1
m = 3
⎡
⎣
⎢
⎢
⇒
M (−1;1)
M (3;5)
⎡
⎣
⎢
⎢
*!
€;!=g1!b:s"r";!.H!FI!.12>!.DE23!

y = −x
*!
€;!=g1!b:trX;!.H!FI!.12>!.DE23!

y = −x + 8
*!

HC#)$=;&(!=VE!FI!$H1!.12>!.DE23!FS3!.C'!R&!

y = −x; y = −x + 8
*!!!!!
Bj=).)k\d*)>%e6lF))
H; L1,1!>$?@3A!.BC3$!

cos3x − 4 cos2x + 8 cos x − 5 = 0
*!!
0; L1,1!0M.!>$?@3A!.BC3$!

log
2
x + 2log
2x
2 ≥ 2
*!!
H; u$?@3A!.BC3$!.?@3A!7?@3A!5g1•!!
!

4cos
3
x −3cosx −4(2cos
2
x −1) + 8cos x −5 = 0
⇔ 4cos
3
x −8cos
2
x + 5cos x −1= 0

⇔ (cosx −1)(2cosx −1)
2
= 0 ⇔
cos x =1
cos x =
1
2
⎡
⎣
⎢
⎢
⎢
⎢
⇔
x = k2π
x = ±
π
3
+ k2π
⎡
⎣
⎢
⎢
⎢
⎢
*!
HC#)$=;&(!=VE!>$?@3A!.BC3$!FI!3A$1J'!R&!

x = k2π, x = ±
π

3
+ k2π,k ∈ !
*!!!
#PD!<•!

cos2x =1+ sin2x ⇔ cos2x −sin2x =1 ⇔ cos(2x +
π
4
) =
1
2
⇔
2x +
π
4
=
π
4
+ k2π
2x +
π
4
= −
π
4
+ k2π
⎡
⎣
⎢
⎢

⎢
⎢
⎢
⎢
⇔
x = kπ
x = −
π
4
+ kπ
⎡
⎣
⎢
⎢
⎢
⎢
,k ∈ !
*!
=VE!>$?@3A!.BC3$!FI!3A$1J'!R&!

x = kπ, x = −
π
4
+ kπ,k ∈ !
*!!
0; y1]D!j1J3•!

x > 0
2x ≠1
⎧

⎨
⎪
⎪
⎩
⎪
⎪
⇔ 0 < x ≠
1
2
*!
`M.!>$?@3A!.BC3$!.?@3A!7?@3A!5g1•!

log
2
x +
2
log
2
2x
≥ 2 ⇔ log
2
x +
2
log
2
x +1
≥ 2
*!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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B4%)#%C#()DE#4$%&2:FG&!
t!
yh.!

t = log
2
x
c!0M.!>$?@3A!.BC3$!.B•!.$&3$•!
!

t +
2
t +1
≥ 2 ⇔
t
2
+ t + 2
t +1
≥ 2 ⇔
t
2
−t
t +1
≥ 0 ⇔
t ≥1
−1< t ≤ 0
⎡
⎣
⎢
⎢

⇔
log
2
x ≥1
−1< log
2
x ≤ 0
⎡
⎣
⎢
⎢
⎢
⇔
x ≥ 2
1
2
< x ≤1
⎡
⎣
⎢
⎢
⎢
⎢
*!
HC#)$=;&(!NV>!3A$1J'!FGH!0M.!>$?@3A!.BC3$!R&!

S =
1
2
;1

⎛
⎝
⎜
⎜
⎜
⎜
⎤
⎦
⎥
⎥
∪ 2;+∞
⎡
⎣
⎢
)
*!!!
!!!!
F; y1]D!j1J3•!

x >1
x < −2
⎡
⎣
⎢
⎢
*!
`M.!>$?@3A!.BC3$!.?@3A!7?@3A!5g1•!
!

x −1

x
>
x −1
x + 2
⇔ (x −1)(
1
x
−
1
x + 2
) > 0 ⇔
2(x −1)
x (x + 2)
> 0 ⇔
x >1
x < −2
⎡
⎣
⎢
⎢
*!
=VE!.V>!3A$1J'!FGH!0M.!>$?@3A!.BC3$!R&!

S = (−∞;−2) ∪ (1;+∞)
*!
Bj=)7)k\d*)>%e6lF!NO3$!.OF$!>$P3!

I =
e
x

e
x
+ 2 e
x
+ 3
dx
0
ln 6
∫
*!
yh.!

t = e
x
+ 3 ⇒ e
x
= t
2
− 3 ⇒ e
x
dx = 2tdt
*!
NH!FI•!
!

I =
2tdt
t
2
−3+ 2t

2
3
∫
=
2tdt
(t −1)(t + 3)
2
3
∫
=
1
2
3
t + 3
+
1
t −1
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
2
3
∫
dt
=

1
2
3ln t + 3 + ln t −1
⎡
⎣
⎢
⎤
⎦
⎥
3
2
=
1
2
(3ln
6
5
+ ln 2)
*!
Bj=)])k\d*)>%e6lF)))
H; LQ1!

z
1
,z
2
R&!$H1!3A$1J'!FGH!>$?@3A!.BC3$!

z
2

− 3z + 5 = 0
*!NC'!>$S3!,%!FGH!()!>$TF!

w = (z
1
2
+ z
2
2
).i + 2−3i
*!)
0; LQ1!U!R&!.V>!$W>!F-F!()!./!3$143!FI!X!F$Y!()!5&!F$1H!$2.!F$%!Z*!NO3$!()!>$S3!.[!FGH!U!5&!
.\3A!FGH!.M.!F,!F-F!>$S3!.[!7I*)
H; NH!FI•!

Δ = 3
2
− 4.5 = −11= 11i
2
⇒ z
1
=
3− 11i
2
,z
2
=
3+ 11i
2
*!

+$1!7I•!

w = (
3− 11i
2
)
2
+ (
3+ 11i
2
)
2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
.i + 2−3i = 2− 4i
*!
Kết$luận:!=VE!>$S3!,%!FGH!‚!0K3A!sƒ*!!
B4c)3F!

z
1
2

+ z
2
2
= (z
1
+ z
2
)
2
− 2z
1
z
2
= 3
2
− 2.5 = −1
*!!
0; #-F!F$Y!()!.$DmF!U!R&!

10003,10010, ,99995
*!#$„3A!RV>!.$&3$!'m.!FM>!()!Fm3A!5g1!()!$a3A!
7SD!0K3A!"………tc!Ff3A!(H1!0K3A!Zc!()!$a3A!FD)1!R&!††††X*!!!!
=VE!.B%3A!U!FI!.M.!F,!3!>$S3!.[!.$%,!'{3•

99995 =10003+ (n −1).7 ⇔ n =12857
*!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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B4%)#%C#()DE#4$%&2:FG&!
ƒ!

€;!N\3A!FGH!.M.!F,!F-F!>$S3!.[!FGH!U!0K3A•

S =
(a
1
+ a
n
).n
2
=
(10003+ 99995).12857
2
= 707122143
*!!
#$„!‡!.\3A!FGH!3!()!$a3A!7SD!FGH!FM>!()!Fm3A!

S =
(a
1
+ a
2
).n
2
*!
oU%)#;@)#<p&1)#q)m)
LQ1!U!R&!.V>!$W>!.M.!F,!F-F!()!./!3$143!A8'!ƒ!F$Y!()!5&!F$1H!$2.!F$%!ˆ*!NO3$!()!>$S3!.[!FGH!U!
5&!.\3A!FGH!.M.!F,!F-F!>$S3!.[!7I*!!
Bj=)_)k\d*)>%e6lF!#$%!$C3$!F$I>!.H'!A1-F!7]D!^*_`#!FI!7-E!_`#!R&!.H'!A1-F!7]D!Fa3$!<H*!LQ1!
bcd!RS3!R?W.!R&!.BD3A!71e'!FGH!F-F!7%a3!`#c^b*!`12.!_d!5Df3A!AIF!5g1!'h.!>$i3A!:^`#;*!
NO3$!.$e!.OF$!j$)1!F$I>!^*_`#!5&!j$%,3A!F-F$!A1YH!$H1!7?k3A!.$i3A!^_!5&!`d*!

!
€;!LQ1!n!R&!.BQ3A!.P'!.H'!A1-F!_`#c!.$~%!A1,!.$12.!.H!FI•!

SO ⊥ (ABC )
*!!
€;!NH!FI•!

AM = AB.sin60
0
= 2a.
3
2
= a 3;SO =
2
3
AM =
2a 3
3
*!
q%!

AN ⊥ (SBC ) ⇒ AN ⊥ SM
*!=C!_d!5‰H!R&!7?k3A!FH%!5‰H!R&!
.BD3A!.DE23!FGH!.H'!A1-F!^_b!343!^_b!FP3!.a1!_*!
^DE!BH•

SA = AM = a 3
*!!
€;!NH'!A1-F!5Df3A!_^n!FI!

SO = SA
2
− AO
2
= 3a
2
−
12a
2
9
=
a 15
3
*!
=C!5VE!

V
S .ABC
=
1
3
SO.S
ABC
=
1
3
.
a 15
3
.

(a 3)
2
. 3
4
=
a
3
5
4
:75 ;*!
€;!NO3$!w:^_r`d;*!
#-F$!"•!^[!wl3A!$J!.BlF!.%a!7m!noEp!FI!A)F!b:…r…r…;c!`:sHr…r…;c!#:Hr…r…;c!

O(0;
a 3
3
;0),A(0;a 3;0),S (0;
a 3
3
;
2a 3
3
)
*!
NH!FI!d!R&!.BD3A!71e'!^b!343!

N (0;
a 3
6
;

a 3
3
)
*!
NH!FI•!!!

SA
! "!
= (0;
2a 3
3
;−
2a 3
3
),BN
! "!!
= (a;
a 3
6
;
a 3
3
),BA
! "!
= (a;a 3;0)
⇒ SA
! "!
,BN
! "!!
⎡

⎣
⎢
⎤
⎦
⎥
= (a
2
;−
2a
2
3
3
;−
2a
2
3
3
), SA
! "!
,BN
! "!!
⎡
⎣
⎢
⎤
⎦
⎥
.BA
! "!
= −a

3
*!
=C!5VE!!
!

d (SA,BN ) =
SA
! "!
,BN
! "!!
⎡
⎣
⎢
⎤
⎦
⎥
.BA
! "!
SA
! "!
,BN
! "!!
⎡
⎣
⎢
⎤
⎦
⎥
=
a

3
a
4
+
4a
4
3
+
4a
4
3
=
a 33
11
!*!
+2.!RDV3•!

V
S .ABC
=
a
3
5
4
;d (SA;BN ) =
a 33
11
*!!
BI94).(!}C3$!j$f3A!A1H3!.$DS3!.D‡•!
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!
NH!FI•!

AN ⊥ (SBC )
343!R„F!3&E!w/3A!Ra1!$C3$!F%1!7-E!R&!^`#!5&!7?k3A!FH%!R&!_d•!
+Š!.1H!^o‹‹`d!.H!FI!`d‹‹:^_c^o;!5C!5VE!
!

d (BN ;SA) = d (BN ;(SA,Sx )) = d(N ;(SA,Sx ))
*!
U~'!FV>!3$V.!F$1!.12.!.B%3A!`%o!7]!()!"ƒ*!!!!
Bj=)-)k\d*)>%e6lF!NB%3A!j$f3A!A1H3!5g1!$J!.BlF!.%a!7m!noEp!F$%!$C3$!0C3$!$&3$!_`#q!FI!
_:"r<rst;c!`:trXrsZ;c!#:"rtrs<;*!NC'!.%a!7m!71e'!q*!=12.!>$?@3A!.BC3$!'h.!>$i3A!:u;!71!vDH!q!5&!
F$TH!7?k3A!.$i3A!n`!5g1!n!R&!A)F!.%a!7m*!
€;!q%!_`#q!R&!$C3$!0C3$!$&3$!343!

AB
! "!!
= DC
! "!!
= (2;3;−4) ⇒
1− x
D
= 2
3− y
D
= 3
−2− z

D
= −4
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⇔
x
D
= −1
y
D
= 0
z
D
= 2
⎧
⎨
⎪
⎪
⎪
⎪
⎩

⎪
⎪
⎪
⎪
⇒ D(−1;0;2)
*!
€;!NH!FI•!

OB
! "!
= (3;5;−7),OD
! "!!
= (−1;0;2)
*!
bh.!>$i3A!:u;!71!vDH!q!5&!F$TH!7?k3A!.$i3A!n`!343!FI!5.>.!R&

n
!
= OB
" !"
,OD
" !""
⎡
⎣
⎢
⎤
⎦
⎥
= (10;1;5)
*!

=C!5VE!

(P ) :10(x +1) +1(y −0) + 5(z −2) = 0 ⇔ (P ) :10x + y + 5z = 0
*!
HC#)$=;&(!=VE!

D(−1;0;2),(P ) :10x + y + 5z = 0
*!!
Bj=),)k\d*)>%e6lF!NB%3A!'h.!>$i3A!5g1!.BlF!.%a!7m!noE!F$%!$C3$!.$%1!_`#q!FI!w1J3!.OF$!0K3A!

2 3
c!>$?@3A!.BC3$!7?k3A!F$x%!`q!R&!

3x − y = 0
*!yz3$!_!R&!$C3$!F$12D!5Df3A!AIF!FGH!`!.B43!
7?k3A!.$i3A!

d : 3x + y = 0
*!NC'!.%a!7m!0)3!7z3$!_c`c#cq!012.!!7z3$!_!FI!$%&3$!7m!w?@3A*!
!
}H1!7?k3A!.$i3A!`q!5&!w!FŒ.!3$HD!.a1!A)F!.%a!7m!n:…r…;!5&!AIF!A1YH!
F$„3A!0K3A!
!

cos(d
1
,d
2
) =
3. 3 −1.1

3+1. 3+1
=
1
2
⇒ BOA
!
= 60
0
*!
€;!LQ1!•!R&!.P'!$C3$!.$%1!_`#qc!.$C!.H'!A1-F!n_#!5Df3A!.a1!_!343!

IAB
!
= BOA
!
= 60
0
*!!
NH!FI•!

S
ABCD
= 4S
IAB
= 2AI .AB sin 60
0
= 3 OA.sin 60
0
( )
. OA.tan 60

0
( )
=
3 3
2
OA
2
*!
N$~%!A1,!.$12.!.H!FI•

3 3
2
OA
2
= 2 3 ⇔ OA
2
=
4
3
*!!
LQ1!_:orE;!5g1!oŽ…!.H!FI!$J!>$?@3A!.BC3$•!

x
2
+ y
2
=
4
3
3x + y = 0

⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⇒ A(
1
3
;−1)
*!
€;!y?k3A!.$i3A!_`!71!vDH!_!5&!5Df3A!AIF!5g1!w!343!FI!>.!R&!

x − 3y−
4
3
= 0
*!
N%a!7m!71e'!`!R&!3A$1J'!FGH!$J!

x − 3y−
4
3
= 0
3x − y = 0

⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⇔
x = −
2
3
y = −2
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⇒ B(−
2
3

;−2)
*!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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B4%)#%C#()DE#4$%&2:FG&!
ˆ!
€;!y?k3A!.$i3A!_•!5Df3A!AIF!5g1!`q!FI!>.!R&!

x + 3y +
2
3
= 0
*!
N%a!7m!71e'!•!R&!3A$1J'!FGH!$J!

3x − y = 0
x + 3y +
2
3
= 0
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪

⎪
⇒ I −
1
2 3
;−
1
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
*!
€;!=C!•!R&!.BD3A!71e'!FGH!_#!5&!`q!343!

C −
2
3
;0
⎛
⎝
⎜
⎜

⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
,D
1
3
;1
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
*!
HC#)$=;&(!=VE!

A(
1
3

;−1),B (−
2
3
;−2),C (−
2
3
;0),D(
1
3
;1)
*!!!!!!
Bj=)`)k\d*)>%e6lF!L1,1!$J!>$?@3A!.BC3$!

(x −1)
3
− 2x( y + 3) + 2y − 2 = 0
(x + y)
2
+ y(1− −2x −1) = x
2
− (−2x −1)
3
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪

⎪
*!
y1]D!j1J3•!

x ≤−
1
2
*!u$?@3A!.BC3$!.$T!$H1!FGH!$J!.?@3A!7?@3A!5g1•!
!

y
2
+ 2xy + y + (−2x −1)
3
= y −2x −1
⇔ y
2
− y −2x −1 + (2x +1)y −(2x +1) −2x −1 = 0
⇔ (y − −2x −1)( y + 2x +1) = 0 ⇔
y = −2x −1
y = −2x −1
⎡
⎣
⎢
⎢
⎢
*!
€;!d2D!

y = −2x −1

.$HE!5&%!>$?@3A!.BC3$!7SD!FGH!$J!.H!7?WF•!

x
3
+ x
2
−5x −5 = 0 ⇔ (x +1)(x
2
−5) = 0 ⇔
x = −1(t / m)
x = − 5(t / m)
x = 5(l )
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⇒
x = −1, y =1
x = − 5, y = 2 5 −1
⎡
⎣
⎢
⎢
⎢
!*!
€;!d2D!

y = −2x −1
.$HE!5&%!>$?@3A!.BC3$!7SD!FGH!$J!.H!7?WF•!
!

(x −1)
3
+ ( −2x −1 +1)
3
= 0 ⇔ −2x −1 +1 = 1− x
⇔ −2x −1 = −x ⇔
x ≤ 0
x
2
= −2x −1
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇔ x = −1 ⇒ y = 1
*!
=VE!$J!>$?@3A!.BC3$!FI!$H1!3A$1J'!R&!

(x; y) = (−1;1);(− 5;2 5 −1)
*!
Bj=)+)k\d*)>%e6lF)#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!.$%,!'{3!

a

2
(
a
c
+1)+ b
2
(
b
c
+1) = 3
*!NC'!A1-!.B9!
3$|!3$M.!FGH!01eD!.$TF

P =
a + c
b
3
+ 2
+
b + c
a
3
+ 2
− 2 a + b + c
*!
^[!wl3A!0M.!7i3A!.$TF!_b!•Lb:wa3A!3A?WF!wMD;!.H!FI•!
!

a + c
b

3
+ 2
=
1
2
a + c −
(a + c)b
3
b
3
+ 2
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
≥
1
2
a + c −
(a + c)b
3
3b
⎡
⎣
⎢
⎢

⎤
⎦
⎥
⎥
=
1
2
a + c −
(a + c)b
2
3
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
*!
N?@3A!./!.H!FI•!

b + c
a
3
+ 2
≥
1
2
b + c −

(b + c )a
2
3
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
*!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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B4%)#%C#()DE#4$%&2:FG&!
Z!
^DE!BH•!
!

a + c
b
3
+ 2
+
b + c
a
3
+ 2
≥
1

2
a + b + 2c −
(a + c)b
2
+ (b + c )a
2
3
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
=
1
2
a + b + 2c −
ab(a + b) + c (a
2
+ b
2
)
3
⎡
⎣
⎢
⎢
⎤

⎦
⎥
⎥
≥
1
2
(a + b + c )
*!
`•1!5C!.$~%!A1,!.$12.!.H!FI•!

c(3−a
2
−b
2
) = a
3
+ b
3
≥ ab(a + b) ⇒ ab(a + b) + c (a
2
+ b
2
) ≤ 3c
*!!
=C!5VE!!
!

P ≥
1
2

(a + b + c )− 2 a + b + c =
( a + b + c −2)
2
2
− 2 ≥−2
*!
qMD!0K3A!7a.!.a1!

a + b + c = 2
a = b =1
a
2
(a + c) + b
2
(b + c ) = 3c
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⇔
a = b =1

c = 2
⎧
⎨
⎪
⎪
⎩
⎪
⎪
*!
=VE!A1-!.B9!3$|!3$M.!FGH!u!0K3A!s<*!!!
oU%)#;@)#<p&1)#q)m))
oU%):O)*\F!#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!F$T3A!'13$!BK3A!
!

a + b
c
3
+ 2
+
b + c
a
3
+ 2
+
c + a
b
3
+ 2
≥ a + b + c −
a

3
+ b
3
+ c
3
3
*!
oU%):O)*.F!#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!.$%,!'{3!

a
3
+ b
3
+ c
3
= 3
*!NC'!A1-!.B9!3$|!3$M.!FGH!
01eD!.$TF!

P =
a + b
c
3
+ 2
+
b + c
a
3
+ 2
+

c + a
b
3
+ 2
− 2(a
2
+ b
2
+ c
2
)
*!
oU%):O)*7F!#$%!Hc0cF!R&!F-F!()!.$/F!w?@3A!.$%,!'{3!

a
2
(
a
c
+1)+ b
2
(
b
c
+1) = 3
*!NC'!A1-!.B9!3$|!
3$M.!FGH!01eD!.$TF!

P = (a + b + c )
2

b + c
a
3
+ 2
+
a + c
b
3
+ 2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
−
9(a + b + c)
2
*!!!!!!
!
!