Đề thi thử và cách giải môn toán thầy đặng thành nam (10)

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

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

m = −2
*!!
=* !>?'!'!7@!:";!7A.!B/B!7A1!.A1!71@'!

x
1
C!7A.!B/B!.1@D!.A1!71@'!

x
2
(E%!B$%!

x
1
2
− x
2
= −1
*!!!!
Bi=).)j\c*)>%d6kF)
E; F1,1!G$HI3J!.K?3$!

log
18
(3
x
+ 2) =
1− x
2+ log
3
2
*!
0; F1,1!G$HI3J!.K?3$!

sin 2x.sin x −
3π
2
⎛
⎝
⎜
⎜
⎜

⎜
⎞
⎠
⎟
⎟
⎟
⎟
= 1+ sin x + tan x
*!!!
Bi=)7)j\c*)>%d6kF!>L3$!.LB$!G$M3!

I = 2x −
1
x
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
ln
2
x dx
1

2
∫
*!
Bi=)l)j\c*)>%d6kF!!
E; >?'!B-B!()!.$/B!EC0CB!(E%!B$%!$E1!G$HI3J!.K?3$!

az
2
+ bz + c = 0;cz
2
+ bz + a +16−16i = 0
:N3!
O;!BP!!3J$1Q'!B$D3J!R&!

1+ 2i
*!
0; #$%!3!R&!()!./!3$143!.$%,!'S3!

C
n
2
+ A
n
2
= n
2
+ 35
*!>?'!()!$A3J!T$U3J!B$VE!W!.K%3J!T$E1!
.K1@3!

x
2
−
2
x
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
n
*!!!!
Bi=)^)j\c*)>%d6kF)#$%!$?3$!$XG!B$Y!3$Z.![\#]*[^\^#^]^!BP!.$@!.LB$!0_3J!"*!F`1!aCbCc!Rd3!
RHe.!R&!.KD3J!71@'!B-B!7%A3!.$f3J![[^C#]C[^]*!>L3$!.$@!.LB$!T$)1!.V!g1Q3!\abc!5&!J1-!.K9!
R<3!3$h.!BiE!T$%,3J!B-B$!J1YE!$E1!7Hj3J!.$f3J!#a!5&![^b*!!
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.V!g1Q3![\#]!BP!.$@!.LB$!0_3J!=*!!
Bi=),)j\c*)>%d6kF)>K%3J!'s.!G$f3J!5<1!.KkB!.%A!7X!lWm!B$%!$?3$!B$Y!3$Z.![\#]!BP!

AC = 2BC
G$HI3J!.K?3$!7Hj3J!B$u%![#!R&!

3x − y− 3 = 0
*!F`1!F!R&!.K`3J!.M'!BiE!.E'!J1-B!
[#]!5&!71@'!

H 3;
2
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
R&!.K/B!.M'!.E'!J1-B![\F*!r12.!G$HI3J!.K?3$!7Hj3J!.$f3J![]*!
Bi=)a)j\c*)>%d6kF)F1,1!$Q!G$HI3J!.K?3$!

x
y +1
x +1
+ y
x +1
y +1
=

x + y + 2xy
2
x
2
+ y
2
−3xy + 5y = 4 x ( 5y −1 − x )
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
(x, y ∈ !)
*!!!
Bi=)+)j\c*)>%d6kF!#$%!WCmCO!R&!B-B!()!.$/B!gHI3J!.$%,!'S3!

x
2
+ 8y
2
+ 8z
2
= 20
*!>?'!J1-!.K9!R<3!

3$h.!BiE!01@D!.$VB!

P =
2x
2
x
2
(4 − yz) + 8
+
y + z
x + y + z +1
−
x
2
+ ( y + z)
2
100
*!
mmm!nLmmm)
M!oV)LpB!)qrV!)estV)uv)/wM)wV)
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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=!
Bi=)\)j.c*)>%d6kF!#$%!$&'!()!

y = x
3
− (m + 2)x
2

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

m = −2
*!!
=* !>?'!'!7@!:";!7A.!B/B!7A1!.A1!71@'!

x
1
C!7A.!B/B!.1@D!.A1!71@'!

x
2
(E%!B$%!

x
1
2
− x
2
= −1
*!!!!
"* v`B!(13$!./!J1,1*!
=* >E!BPw!

y ' = 3x
2
− 2(m + 2)x + 2m +1; y ' = 0 ⇔
x = 1

x =
2m +1
3
⎡
⎣
⎢
⎢
⎢
⎢
*!
x@!:";!BP!=!B/B!.K9!T$1!m^!BP!=!3J$1Q'!G$M3!01Q.!

⇔
2m +1
3
≠ 1 ⇔ m ≠1
*!
y!b2D!

m >1 ⇒1<
2m +1
3
⇒ x
1
= 1;x
2
=
2m +1
3
*!

z4D!BdD!.K{!.$&3$w!

1
2
−
2m +1
3
= −1 ⇔ m =
5
2
(t / m)
*!
y!b2D!

m <1 ⇒
2m +1
3
<1 ⇒ x
1
=
2m +1
3
,x
2
= 1
*!
z4D!BdD!.K{!.$&3$w!

2m +1
3

⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
2
−1 = −1 ⇔
2m +1
3
= 0 ⇔ m = −
1
2
(t / m)
*!
rZm!

m = −
1
2
;m =
5
2
R&!J1-!.K9!Bd3!.?'*!!!!!

Bi=).)j\c*)>%d6kF)
E; F1,1!G$HI3J!.K?3$!

log
18
(3
x
+ 2) =
1− x
2+ log
3
2
*!
0; F1,1!G$HI3J!.K?3$!

sin 2x.sin x −
3π
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟

= 1+ sin x + tan x
*!!!
E; c$HI3J!.K?3$!.HI3J!7HI3J!5<1w!

log
18
(3
x
+ 2) =
1− x
log
3
18
⇔ log
18
(3
x
+ 2) = (1− x )log
18
3
⇔ log
18
(3
x
+ 2) = log
18
3
1−x
⇔ 3
x

+ 2 = 3
1−x
=
3
3
x
⇔ 3
2x
+ 2.3
x
−3 = 0 ⇔
3
x
= 1
3
x
= −3(l )
⎡
⎣
⎢
⎢
⎢
⇔ x = 0
*!
rZm!G$HI3J!.K?3$!BP!3J$1Q'!gDm!3$h.!

x = 0
*!!
0; x1|D!T1Q3w!

cos x ≠ 0 ⇔ x ≠
π
2
+ kπ,k ∈ !
*!
c$HI3J!.K?3$!.HI3J!7HI3J!5<1w!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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q!

sin2x.cosx =1+sin x +
sin x
cos x
⇔ 2sin x.cos
3
x = cosx +sin x cos x + sin x
⇔ sin x cos x 2cos
2
x −1
( )
= sin x + cos x
⇔ sin 2x cosx −sin x
( )
cos x +sin x
( )
= 2 sin x + cosx
( )
⇔ sin x + cos x
( )

sin2x cos x −sin x
( )
− 2
( )
= 0
⇔
sin x + cos x = 0 (1)
sin2x cos x −sin x
( )
− 2 = 0 (2)
⎡
⎣
⎢
⎢
⎢
*!
y;!>E!BP!

(1) ⇔ sin x = −cos x ⇔ tan x = −1 ⇔ x = −
π
4
+ kπ
*!
y;!>E!BP!

VT
(2)
≤ sin 2x . sin x − cos x − 2≤ 2 −2 < 0
:5U!3J$1Q';*!
5Zm!G$HI3J!.K?3$!BP!3J$1Q'!R&!

x = −
π
4
+ kπ,k ∈ !
*!!!!!
Bi=)7)j\c*)>%d6kF!>L3$!.LB$!G$M3!

I = 2x −
1
x
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
ln
2
x dx
1
2
∫
*!

>E!BP!

I = 2 x ln
2
x
1
2
∫
K
! "### $###
−
ln
2
x
x
dx
1
2
∫
= 2K − ln
2
xd(ln x )
1
2
∫
= 2K −
ln
3
x
3

2
1
= 2K −
ln
3
2
3
*!
xs.!

u = ln
2
x
dv = xdx
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒
du =
2ln x
x
dx
v =
x
2
2

⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
*!
}Dm!KE!

K =
x
2
ln
2
x
2
2
1
− x ln x dx
1
2
∫
M

! "#### $####
= 2ln
2
2−M
*!
xs.!

u = ln x
dv = xdx
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒
du =
dx
x
v =
x
2
2
⎧
⎨
⎪
⎪
⎪
⎪

⎪
⎩
⎪
⎪
⎪
⎪
⎪
⇒ M =
x
2
ln x
2
2
1
−
1
2
x dx
1
2
∫
= 2ln 2−
1
4
x
2
2
1
= 2ln 2−
3

4
*!
r?!5Zm!

I = 2 2ln
2
2−(2ln 2−
3
4
)
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
−
ln
3
2
3
= −
ln
3

2
3
+ 4 ln
2
2−4ln 2+
3
2
*!!!
Bi=)l)j\c*)>%d6kF!!
B; >?'!B-B!()!.$/B!EC0CB!(E%!B$%!$E1!G$HI3J!.K?3$!

az
2
+ bz + c = 0;cz
2
+ bz + a +16−16i = 0
:N3!
O;!BP!!3J$1Q'!B$D3J!R&!

1+ 2i
*!
g; #$%!3!R&!()!./!3$143!.$%,!'S3!

C
n
2
+ A
n
2
= n

2
+ 35
*!>?'!()!$A3J!T$U3J!B$VE!W!.K%3J!T$E1!
.K1@3!

x
2
−
2
x
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
n
*!!!!
E;!>$~%!J1,!.$12.!G$HI3J!.K?3$!

az
2
+ bz + c = 0

BP!3J$1Q'!:"y=1;!T$1!!!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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•!
!

a(1+ 2i )
2
+ b(1+ 2i) + c = 0
⇔ −3a + b + c + (4a + 2b)i = 0
⇔
−3a + b + c = 0
4a + 2b = 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
(1)
*!
>HI3J!./!G$HI3J!.K?3$!

cz
2
+ bz + a +16 −16i = 0
!BP!3J$1Q'!:"y=1;!T$1!!

c(1+ 2i )
2
+ b(1+ 2i) + a +16−16i = 0
⇔ c(−3+ 4i )+ b + 2bi + a +16−16i = 0
⇔ (a + b − 3c +16) + 2(b + 2c −8)i = 0
⇔
a + b − 3c +16 = 0
b + 2c − 8 = 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
(2)
*!
>€!:";!5&!:=;!(Dm!KE!

(a;b;c) = (1;−2;5)
*!
!
0;!>$~%!J1,!.$12.!.E!BPw!

n(n −1)
2
+ n(n −1) = n
2
+ 35 ⇔ n
2

−3n −70 = 0 ⇔
n =10(t / m)
n = −7(l )
⎡
⎣
⎢
⎢
*!
+$1!7P!

x
2
−
2
x
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
10
= C

10
k
x
20−5k
(−2)
k
k=0
10
∑
*!
})!$A3J!T$U3J!B$VE!W!T$1!

20−5k = 0 ⇔ k = 4
*!!!
rZm!()!$A3J!Bd3!.?'!R&!

C
10
4
(−2)
4
*!!
Bi=)^)j\c*)>%d6kF)#$%!$?3$!$XG!B$Y!3$Z.![\#]*[^\^#^]^!BP!.$@!.LB$!0_3J!"*!F`1!aCbCc!Rd3!
RHe.!R&!.KD3J!71@'!B-B!7%A3!.$f3J![[^C#]C[^]*!>L3$!.$@!.LB$!T$)1!.V!g1Q3!\abc!5&!J1-!.K9!
R<3!3$h.!BiE!T$%,3J!B-B$!J1YE!$E1!7Hj3J!.$f3J!#a!5&![^b*!!
!
y;!•ZG!$Q!.KkB!.%A!7X!BP![:onono;C!\:Enono;C!#:En0no;C!]:on0no;C!
[^:ononB;C!\^:EnonB;C!#^:En0nB;C]^:on0nB;*!
+$1!7P!

M (0;0;
c
2
),N (
a
2
;b;0),P(0;
b
2
;
c
2
)
*!
>$~%!J1,!.$12.!.E!BPw!

abc = 1
*!!
y;!>E!BP!

V
BMNP
=
1
6
BM
! "!!
,BN
! "!!
⎡

⎣
⎢
⎤
⎦
⎥
.BP
! "!
=
5abc
48
=
5
48
*!!!!
y;!>L3$!7HeBw!

d (CM ; A' N ) =
CM
! "!!
,A' N
! "!!!
⎡
⎣
⎢
⎤
⎦
⎥
.MA'
! "!!!
CM

! "!!
,A' N
! "!!!
⎡
⎣
⎢
⎤
⎦
⎥
=
abc
4b
2
c
2
+ 4a
2
b
2
+ 9a
2
c
2
=
1
4
a
2
+
4

c
2
+
9
b
2
*!
}‚!gk3J!0h.!7f3J!.$VB![a!ƒFa!.E!BPw!
!

4
a
2
+
4
c
2
+
9
b
2
≥ 3
4
a
2
.
4
c
2
.

9
b
2
3
= 3 144
3
⇒ d (CM ; A' N ) ≤
1
3 144
3
*!
]hD!0_3J!7A.!.A1!

a
2
=
c
2
=
b
3
=
1
12
3
*!rZm!J1-!.K9!R<3!3$h.!BiE!T$%,3J!B-B$!J1YE!#a!5&![^b!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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„!

0_3J!

1
3 144
3
*!!
B4b)3(!\&1!.%-3!J1,!.$12.!$?3$!$XG!B$Y!3$Z.C!$?3$!RZG!G$HI3J!32D!T$P!.12G!BZ3!BP!.$@!J…3!.KkB!
.%A!7X!7@!W‚!R†*!
Bi=)-)j\c*)>%d6kF!>K%3J!T$U3J!J1E3!5<1!$Q!.KkB!.%A!7X!lWmO!B$%!0E!71@'![:"nono;C!\:onp=no;C!
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.V!g1Q3![\#]!BP!.$@!.LB$!0_3J!=*!!
>E!BPw!

AB
! "!!
= (−1;−2;0),AC
! "!!
= (0;−2;3) ⇒ AB
! "!!
,AC
! "!!
⎡
⎣
⎢
⎤
⎦
⎥
= (−6;3;2)
*!
as.!G$f3J!71!tDE!0E!71@'![C\C#!3$Z3!!

AB
! "!!
,AC
! "!!
⎡
⎣
⎢
⎤
⎦
⎥
= (−6;3;2)
R&'!5uB!.I!G$-G!.Dm23!343!BP!
G$HI3J!.K?3$!R&!

−6(x −1)+ 3y + 2z = 0 ⇔ 6x − 3y − 2z −6 = 0
*!
y;!F`1!]:ononE;!5<1!E‡o!.E!BP!

S
ABC
=
1
2
AB
! "!!
,AC
! "!!
⎡
⎣

⎢
⎤
⎦
⎥
=
7
2
;d (D;(ABC )) =
−2a − 6
7
*!
>E!BPw!

V
ABCD
=
1
3
S
ABC
.d(D;(ABC )) =
1
3
.
7
2
.
−2a − 6
7
= 2 ⇔

a = 3(t / m)
a = −9(l )
⎡
⎣
⎢
⎢
*!
rZm!71@'!Bd3!.?'!R&!]:ononq;*!!!!!
Bi=),)j\c*)>%d6kF)>K%3J!'s.!G$f3J!5<1!.KkB!.%A!7X!lWm!B$%!$?3$!B$Y!3$Z.![\#]!BP!

AC = 2BC
G$HI3J!.K?3$!7Hj3J!B$u%![#!R&!

3x − y − 3 = 0
*!F`1!F!R&!.K`3J!.M'!BiE!.E'!J1-B!
[#]!5&!71@'!

H 3;
2
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟

⎟
⎟
R&!.K/B!.M'!.E'!J1-B![\F*!r12.!G$HI3J!.K?3$!7Hj3J!.$f3J![]*!
#$V3J!'13$!7HeB!v!.$DXB!7%A3!#]!5&!

CH =
2
3
CD
*!!
>E!BP!

AC = 2BC ⇒ CD = 3AD ⇒ ACD
!
= 30
0
*!
F`1!:En0;!R&!5uB!.I!G$-G!.Dm23!BiE!#]!.E!BPw!
!

3a −b
2 a
2
+ b
2
=
3
2
⇔ 3(a
2

+ b
2
) = ( 3a − b)
2
⇔ 2b
2
+ 2 3ab = 0 ⇔ b(b + 3a) = 0 ⇔
b = 0
b = − 3a
⎡
⎣
⎢
⎢
⎢
*!
y;!b2D!

b = 0 ⇒ CD : x − 3 = 0
*!!
>%A!7X!71@'!#!R&!3J$1Q'!BiE!G$HI3J!.K?3$!

x −3 = 0
3x − y − 3 = 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪

⇔
x = 3
y = 2 3
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ C(3;2 3)
*!
>E!BP!

CD
! "!!
=
3
2
CH
! "!!
=
3
2
0;−
4
3
⎛
⎝
⎜

⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
= 0;−2 3
( )
⇒ D(3;0)
*!
xHj3J!.$f3J![]!71!tDE!]!5&!5DU3J!JPB!5<1!#]!BP!G$HI3J!.K?3$w!

y = 0
*!!!!
y;!b2D!

b = − 3a ⇒ CD : x − 3y −1 = 0
*!
>%A!7X!71@'!#!R&!3J$1Q'!BiE!$Q!G$HI3J!.K?3$!

x − 3y −1= 0
3x − y − 3 = 0
⎧
⎨
⎪
⎪
⎪

⎩
⎪
⎪
⎪
⇔
x = 1
y = 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ C(1;0)
*!
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&)!
6!
Ta!có!

CD
! "!!
=
3
2
CH
! "!!
= 3; 3

( )
⇒ D(4; 3)
.!!
Đường!thẳng!AD!đi!qua!điểm!D!và!vuông!góc!với!CD!có!phương!trình:

3x + y −5 3 = 0
.!
HC#)$=;&(!Vậy!có!hai!đường!thẳng!thoả!mãn!yêu!cầu!bài!toán!là

3x + y −5 3 = 0; y = 0
.!!
BI=)J)KLM*)>%N6OF)Giải!hệ!phương!trình!

x
y +1
x +1
+ y
x +1
y +1
=
x + y + 2xy
2
x
2
+ y
2
−3xy + 5y = 4 x ( 5y −1 − x )
⎧
⎨
⎪

⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
(x, y ∈ !)
.!!!
Điều!kiện:!

x ≥ 0; y ≥
1
5
.!!
Sử!dụng!bất!đẳng!thức!AM!–GM!ta!có:!

x
y +1
x +1
+ y
x +1
y +1
≤
1
2
x
y +1
x +1

+1
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
+ y
x +1
y +1
+1
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎡

⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
=
x + y + 2
2
x
x +1
+
y
y +1
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
=

x + y + 2
2
.
x + y + 2xy
(x +1)(y +1)
=
x + y + 2
2(xy + x + y +1)
.(x + y + 2xy)
.!
Dấu!bằng!xảy!ra!kh!và!chỉ!khi!

x = y
.!!
Kết!hợp!với!phương!trình!thứ!nhất!của!hệ!ta!có:!

x + y + 2xy
2
≤
x + y + 2
2(xy + x + y +1)
.(x + y + 2xy) ⇔
x + y + 2
xy + x + y +1
≥1 ⇔ xy ≤1 (1)
.!
Dấu!bằng!xảy!ra!khi!và!chỉ!khi!

xy = 1
.!!

Phương!trình!thứ!hai!của!hệ!tương!đương!với:!

(x − y)
2
− xy + 5y = 4 x(5y −1) − 4x
⇔ (x − y)
2
− xy + 4x + 5y − 4 x(5y −1) = 0
⇔ (x − y)
2
− xy +1+ 4x − 4 x(5y − 1) +5y −1
( )
= 0
⇔ (x − y)
2
− xy +1+(2 x − 5y −1)
2
= 0
⇔ (x − y)
2
+ (2 x − 5y −1)
2
= xy −1
.!
Suy!ra!

xy −1≥ 0 ⇔ xy ≥1 (2)
.!!
Từ!(1)!và!(2)!suy!ra!

xy = 1
.!Vì!vậy!

x = y
xy =1
(x − y)
2
+ (2 x − 5y −1)
2
= 0
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⇔
x = 1
y = 1
⎧
⎨
⎪
⎪
⎩
⎪

⎪
(thử!lại!thấy!thoả!
mãn).!
Vậy!hệ!phương!trình!có!nghiệm!duy!nhất!

(x; y) = (1;1)
.!!!!!!
BI=)+)KLM*)>%N6OF!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn!

x
2
+ 8y
2
+ 8z
2
= 20
.!Tìm!giá!trị!lớn!
nhất!của!biểu!thức!

P =
2x
2
x
2
(4 − yz) + 8
+
y + z
x + y + z +1
−
x

2
+ ( y + z)
2
100
.!
Lời$giải:$
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&)!
7!
Sử!dụng!bất!đẳng!thức!Cauchy!–!Schwarz!ta!có:!
!

x
2
+ ( y + z)
2
≥
1
2
x + y + z
( )
2
.!!
Sử!dụng!bất!đẳng!thức!AM!–GM!ta!có:!
!

2x
2
4x

2
+ 8− x
2
yz
=
x
2x +
4
x
−
xyz
2
=
x
x + x +
4
x
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
−

xyz
2
≤
x
x + 4 −
xyz
2
.!!
Sử!dụng!bất!đẳng!thức!Cauchy!–!Schwarz!ta!có:!
!

x.yz + 2.y + 2.z + 2.1
( )
2
≤ (x
2
+12)( y
2
z
2
+ y
2
+ z
2
+1)
= x
2
+12
( )
y

2
+1
( )
z
2
+1
( )
=
1
64
x
2
+12
( )
8y
2
+ 8
( )
8z
2
+ 8
( )
≤
1
64
x
2
+ 8y
2
+ 8z

2
+ 28
3
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
3
= 64
⇒ xyz + 2y + 2z ≤ 6 ⇒
x
x + 4 −
xyz
2
≤
x
x + y + z +1
.!
Vì!vậy!

P ≤

x + y + z
x + y + z +1
−
(x + y + z)
2
200
.!Đặt!

t = x + y + z > 0
ta!có!

P ≤
t
t +1
−
t
2
200
.!
Xét!hàm!số!

f (t) =
t
t +1
−
t
2
200
!ta!có!

f '(t) =
1
(t +1)
2
−
t
100
; f '(t) = 0 ⇔ t(t +1)
2
= 100 ⇔ t = 4
.!
Vì!f’(t)!đổi!dấu!từ!dương!sang!âm!khi!đi!qua!t=4!nên!

P ≤ f (t ) ≤ f (4) =
18
25
.!
Dấu!bằng!đạt!tại!

x = 2; y = z = 1
.!Vậy!giá!trị!lớn!nhất!của!P!bằng!18/25.!!
BP94).()
Sử dụng bất đẳng thức AM-GM cho hai số dương ta được:

x
2
yz ≤ x
2
.
y

2
+ z
2
2
=
5
4
x
2
−
x
4
16
;
2xy + 2xz ≤
x
2
+ 4 y
2
2
+
x
2
+ 4z
2
2
= x
2
+ 2.
20− x

2
8
=
3
4
x
2
+ 5;
2x ≤
x
2
2
+ 2 ≤ (
x
4
16
+1) + 2 =
x
4
16
+ 3

Cộng theo vế ba bất đẳng thức cùng chiều ta thu được:

x
2
yz + 2xy + 2xz + 2x ≤ 2x
2
+ 8
.

Suy ra:

x
2
(4 − yz) + 8 = 2x(x + y + z +1)+ 2x
2
+ 8−(x
2
yz + 2xy + 2xz + 2x )
≥ 2x(x + y + z +1)
.
Mặt khác, theo bất đẳng thức Cauchy –Schwarz ta có:

x
2
+ ( y + z)
2
≥
(x + y + z)
2
2
.
Từ đó ta có được:

P ≤
x + y + z
x + y + z +1
−
(x + y + z)
2

200
.!Ta!có!kết!quả!tương!tự!trên.
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&)!
8!
Q4;&)RS#F))
+)!Mấu!chốt!bài!toán!tìm!được!điểm!rơi!

x = 2; y = z = 1
.!
+)!Đánh!giá!

xyz + 2 y + 2z ≤ 6
.!
TU%)#;@)#<V&1)#W)X!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn!

x
2
+ y
2
+ z
2
= 26
.!Tìm!giá!trị!lớn!
nhất!của!biểu!thức!

P = xyz + 32 y + 45z
.!
Đ/S:!

P
max
= P (1;3;4) = 288
.!!!!!!!!!!