Đề thi thử và cách giải môn toán thầy đặng thành nam (20)

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0oR%d%pGiC%SAj=qY%#$%!$&'!()!

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

m = 2
*!!
=* >?'!'!7@!$&'!()!:";!AB!A/A!.1@C!D&!

y
CT
.$%,!'E3!

y
CT
=
1
2
*!!!
0oR%G%pdiC%SAj=qY%
F; G1,1!H$IJ3K!.L?3$!

cos
2
x + cos
2
10x = 1+ 3 cos11x.sin 9x
*!!
0; #$%!()!H$MA!

z = 1+ i
*!>N3$!'O7C3!APF!()!H$MA!

w =1+ z + z
3
+ z
4
*!!!
0oR%L%pCic%SAj=qY!G1,1!H$IJ3K!.L?3$!

4
x

− 9
x
= 5(ln x −1)(3
x
− 2
x
)
*!!!
0oR%r%pdiC%SAj=qY!G1,1!$Q!H$IJ3K!.L?3$!

(x + y)(
1
xy
+ 3) =
6(x
2
+ y
2
) + 4
2(x
2
+ y
2
)
4− x
2
− y
2
= 2 2xy + 2− x
2

− y
2
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
(x, y ∈ !)
*!!
0oR%c%pdiC%SAj=qY!>N3$!.NA$!H$R3!

I = (x +1) −x
2
−2x + 3dx
0
1
∫
*!
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O,SO ⊥ (ABCD )
!

Z!

AB = a,BC = a 3
!5&!

SC = a 6
*!G[1!\!D&!71@'!.$C]A!7%^3!S#!.$%,!'E3!

SC = 3IC
*!>N3$!.$@!
.NA$!_$)1!A$BH!S*TU#V!5&!_$%,3K!A-A$!K1XF!$F1!7I`3K!.$a3K!T\!5&!SU*!!!
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(P ) : x − 2y + 2z −1= 0
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m!D&!.LC3K!71@'!A^3$!#VZ!o!D&!$?3$!A$12C!5CO3K!KBA!APF!V!.L43!Tm!5&!p!D&!.LC3K!71@'!To*!
G1,!(q!H$IJ3K!.L?3$!7I`3K!.Lr3!3K%^1!.12H!.F'!K1-A!\mp!D&!

(C ) :(x −
5
2
)
2
+ ( y −
9
2
)
2

=
5
2
!5&!
7s3$!V!AB!$%&3$!7]!3KCW43!3t'!.L43!7I`3K!.$a3K!

x − 2y = 0
*!>?'!.%^!7]!A-A!7s3$!UZV*!
0oR%D%pCic%SAj=qY%>?'!()!$^3K!A$MF!

x
10
!.L%3K!_$F1!.L1@3!

(
nx
5
−
1
x
2
)
n
!012.!3!D&!()!./!3$143!.$%,!
'E3!

n
2
+C
n

2
= 145 (n ≥ 2)
*!
0oR%dC%pdiC%SAj=qY!#$%!eZWZf!D&!A-A!()!.$/A!uIJ3K!.$%,!'E3!

z = min x, y,z
{ }
*!>?'!K1-!.L9!3$v!
3$w.!APF!01@C!.$MA!

P =
x
y + z
+
y
z + x
+
3
2
.
z
x + y
−
z
z
2
+ xy + yz + zx
*!
sss"t+sss%
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03A%?AW?6%X<?3@A8IOYZ8!
=!
,"u;%+v0"%wx;"%k.y;%)$,%$;%
0oR%d%pGiC%SAj=qY%#$%!$&'!()!

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

m = 2
*!!
=* >?'!'!7@!$&'!()!:";!AB!A/A!.1@C!D&!

y
CT
.$%,!'E3!

y
CT
=

1
2
*!!!
"* o[A!(13$!./!K1,1*!
=* >F!ABx!

y ' = 2x
3
− 2m
2
x;y ' = 0 ⇔
x = 0
x
2
= m
2
⎡
⎣
⎢
⎢
*!
y;!p2C!

m = 0 ⇒ y ' = 2x
3
; y ' = 0 ⇔ x = 0
Z!$&'!()!7^.!A/A!.1@C!.^1!

x = 0; y
CT

= 1≠
1
2
:343!D%^1;*!
y;!p2C!

m ≠ 0
!_$1!7B!$&'!()!7^.!A/A!.1@C!.^1!

±m; y
CT
= y(±m) = 1−
m
4
2
*!
>F!AB!H$IJ3K!.L?3$x!

1−
m
4
2
=
1
2
⇔ m
4
= 1 ⇔ m = −1;m = 1
*!
!W?%@RP86!iYW!

m = −1;m =1
D&!K1-!.L9!Aj3!.?'*!!!!!!!
0oR%G%pdiC%SAj=qY%
F; G1,1!H$IJ3K!.L?3$!

cos
2
x + cos
2
10x = 1+ 3 cos11x.sin 9x
*!!
0; #$%!()!H$MA!

z = 1+ i
*!>N3$!'O7C3!APF!()!H$MA!

w =1+ z + z
3
+ z
4
*!!!
F; l$IJ3K!.L?3$!.IJ3K!7IJ3K!5<1x!

1+ cos2x +1+ cos20x
2
= 1+ 3 cos11x.sin9x
⇔
cos2x + cos20x
2

= 3 cos11x.sin9x
⇔ cos11x.cos9x = 3 cos11x.sin9x
⇔
cos11x = 0
tan9x =
1
3
⎡
⎣
⎢
⎢
⎢
⎢
⇔
x =
π
22
+ k
π
11
x =
π
54
+ k
π
9
⎡
⎣
⎢
⎢

⎢
⎢
⎢
⎢
,k ∈ !
*!
!W?%@RP86!iYW!3K$1Q'!APF!H$IJ3K!.L?3$!D&!

x =
π
22
+ k
π
11
;x =
π
54
+ k
π
9
,k ∈ !
*!!
0; >F!ABx!

w =1+ (1+ i)+ (1+ i )
3
+ (1+ i)
4
= −4+ 3i ⇒ w = 5
*!!

0oR%L%pCic%SAj=qY!G1,1!H$IJ3K!.L?3$!

4
x
− 9
x
= 5(ln x −1)(3
x
− 2
x
)
*!!!
z1{C!_1Q3x!

x > 0
*!
l$IJ3K!.L?3$!.IJ3K!7IJ3K!5<1x!
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03A%?AW?6%X<?3@A8IOYZ8!
|!
!

(2
x
−3
x
)(2
x
+ 3

x
) = 5(ln x −1)(3
x
− 2
x
)
⇔ (2
x
−3
x
)(2
x
+ 3
x
+ 5ln x −5) = 0
⇔
2
x
= 3
x
2
x
+ 3
x
+ 5ln x −5 = 0
⎡
⎣
⎢
⎢
⎢

⇔
x = 0(l )
x =1(t / m)
⎡
⎣
⎢
⎢
*!
!W?%@RP86!l$IJ3K!.L?3$!AB!3K$1Q'!uCW!3$w.!

x =1
*!!
0oR%r%pdiC%SAj=qY!G1,1!$Q!H$IJ3K!.L?3$!

(x + y)(
1
xy
+ 3) =
6(x
2
+ y
2
) + 4
2(x
2
+ y
2
)
4− x
2

− y
2
= 2 2xy + 2− x
2
− y
2
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
(x, y ∈ !)
*!!
y;!z1{C!_1Q3x!

xy > 0,x
2
+ y
2
≤ 2
*!
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y;!l$IJ3K!.L?3$!7jC!APF!$Q!.IJ3K!7IJ3K!5<1x!

3(x + y − 2(x
2
+ y
2
)) +
1
x
+
1
y
−
4
2(x
2
+ y
2
)
= 0
⇔ 3(x + y − 2(x
2
+ y
2
)) +
2(x
2
+ y
2
)(x + y)−4xy

xy 2(x
2
+ y
2
)
= 0
( 2(x
2
+ y
2
) −x − y) −3+
2(x
2
+ y
2
)(x + y)−4xy
xy 2(x
2
+ y
2
)( 2(x
2
+ y
2
) −x − y)
⎛
⎝
⎜
⎜
⎜

⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
⎟
= 0
⇔ ( 2(x
2
+ y
2
) −x − y) A−3
( )
= 0 (*)
*!
>L%3K!7B!!
!

A =
( 2(x
2
+ y
2
)(x + y)−4xy)( 2(x
2
+ y

2
) + x + y)
xy 2(x
2
+ y
2
)(x − y)
2
=
( 2(x
2
+ y
2
)(x + y)−4xy)( 2(x
2
+ y
2
) + x + y)
xy 2(x
2
+ y
2
)(x − y)
2
=
2(x + y) + 2(x
2
+ y
2
)

xy 2(x
2
+ y
2
)
*!
V%!

xy ≤
x
2
+ y
2
2
≤1
*!SCW!LFx!
!

2(x + y) + 2(x
2
+ y
2
)
xy 2(x
2
+ y
2
)
≥
3(x + y)

xy 2(x
2
+ y
2
)
≥
6
2xy(x
2
+ y
2
)
≥
6
2.1.2
= 3
*!
i?!5YW!

(*) ⇔
x = y = 1
x = y
⎡
⎣
⎢
⎢
*!
y;!i<1!

x = y =1

.$FW!5&%!H$IJ3K!.L?3$!.$M!$F1!APF!$Q!_$O3K!.$%,!'E3*!
y;!i<1!

x = y
.$FW!5&%!H$IJ3K!.L?3$!.$M!$F1!APF!$Q!.F!7I}Ax!
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03A%?AW?6%X<?3@A8IOYZ8!
~!
!

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

1
2
*!
!W?%@RP86!oQ!H$IJ3K!.L?3$!AB!3K$1Q'!uCW!3$w.!

(x; y) = (
1
2
;
1
2
)
*!!!
wz83%@RP86!l$•H!0123!7€1!H$IJ3K!.L?3$!7jC!_$-!$1QC!nC,Z!7B!D&!7b.!3KFW!3$R3!.q!

2(x
2
+ y
2
) − x − y
!A$MF!.$&3$!H$j3!(FC!D143!$}Hx!

(x − y)
2
!LF!3K%&1!A$%!.$CY3!D}1!0123!7€1!
H$IJ3K!.L?3$*!
w:A%?PU%?Q{89%?|%%
w:A%O^%CdY%G1,1!$Q!H$IJ3K!.L?3$!

(x + y)(1+

7
20xy
) =
10(x
2
+ y
2
) + 7
5 2(x
2
+ y
2
)
2− x
2
− y
2
= 2 2xy −1
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪

⎪
*!z•(x!

(x; y) = (
1
2
;
1
2
)
*!!!!!
w:A%O^%CGY!G1,1!$Q!H$IJ3K!.L?3$!

(2− x )(2− y) +
x
2
+ y
2
2
= 2
2− x
2
− y
2
+ 2 2xy = 4− x
2
− y
2
⎧
⎨

⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
*!z•(x!!

(x; y) = (
1
2
;
1
2
)
*!!!!!
HD:!l$IJ3K!.L?3$!7jC!APF!$Q!.IJ3K!7IJ3K!5<1x!
!

(2− x )(2− y) = 2−
x
2
+ y
2
2
⎛
⎝

⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
⎟
2
⇔ (x − y)
2
2
2(x
2
+ y
2
) + x + y
−
1
2
⎡
⎣
⎢
⎢
⎢

⎤
⎦
⎥
⎥
⎥
= 0 ⇔ x = y
*!
V%!

2
2(x
2
+ y
2
) + x + y
−
1
2
≥
2
2+ 2
−
1
2
= 0
*!!!!
0oR%c%pdiC%SAj=qY!>N3$!.NA$!H$R3!

I = (x +1) −x
2

−2x + 3dx
0
1
∫
*!
zb.!

t = 3− x
2
− 2x ⇒ t
2
= 3− x
2
− 2x ⇒ 2tdt = (−2x −2)dx = −2(x +1)dx
*!
>F!ABx!

I = t.(−t dt )
3
0
∫
= t
2
dt
0
3
∫
=
t
3

3
3
0
= 3
*!!!
0oR%F%pdiC%SAj=qY%%#$%!$?3$!A$BH!S*TU#V!AB!7-W!TU#V!D&!$?3$!A$X!3$Y.!.R'!

O,SO ⊥ (ABCD )
!
Z!

AB = a,BC = a 3
!5&!

SC = a 6
*!G[1!\!D&!71@'!.$C]A!7%^3!S#!.$%,!'E3!

SC = 3IC
*!>N3$!.$@!
.NA$!_$)1!A$BH!S*TU#V!5&!_$%,3K!A-A$!K1XF!$F1!7I`3K!.$a3K!T\!5&!SU*!!!
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">?@A8B6%CDEF%GFF%GCG%% )H89%IJ%83K=%L%3MN%OA83%83P8%QR%STA%3MN%U3V%%%
03A%?AW?6%X<?3@A8IOYZ8!
‚!
!
>F!ABx!

OC =
AC
2

=
a
2
+ 3a
2
2
= a
*!
>F'!K1-A!5CO3K!Sd#!AB!x!

SO = SC
2
−OC
2
= 6a
2
− a
2
= a 5
*!!!
iYW!

V
S . ABCD
=
1
3
SO.S
ABCD
=

1
3
.a 5.a.a 3 =
a
3
15
3
:75 ;*!
y;!+ƒ!\m!(%3K!(%3K!5<1!SU!A„.!U#!.^1!m*!
_ƒ!\o!(%3K!(%3K!5<1!Sd!A„.!T#!.^1!oZ!.F!ABx!

IH
SO
=
CH
CO
=
CI
CS
=
1
3
,IH =
1
3
SO =
a 5
3
*!!
>F!ABx!SU••\m!343!SU••:T\m;!u%!7Bx!

d (SB; AI ) = d (SB;(AIM )) = d (B;(AIM )) = 2d (C ;(AIM )) = 2.
CA
HA
d (H ;(AIM )) =
12
5
d (H ;(AIM ))
*!!
y;!d…!5CO3K!KBA!5<1!Tm!.^1!…Z!o+!5CO3K!KBA!5<1!\…!.^1!+x

HK ⊥ (AIM ) ⇒ HK = d(H ;(AIM ))
*!
>F'!K1-A!Tm#!AB!

S
AMC
=
1
2
S
ABC
=
1
2
.
1
2
.a.a 3 =
a

2
3
4
,AM = a
2
+
4a
2
3
=
a 21
3
*!
SCW!LFx!

d (C ;AM ) =
2S
AMC
AM
=
2.
a
2
3
4
a 21
3
=
3a 7
14

!5&!

HE =
5
6
d (C ;AM ) =
5
6
.
3a 7
14
=
5a 7
28
*!!
SCW!LFx!!
!

1
HK
2
=
1
IH
2
+
1
HE
2
=

9
5a
2
+
112
25a
2
=
157
25a
2
⇒ HK =
5a
157
*!!
!W?%@RP86!iYW!

d (SB; AI ) =
12
5
HK =
12a
157
*!!
0oR%E%pdiC% SAj=qY! >L%3K! 'b.! H$a3K! 5<1! .LcA! .%^! 7]! deWf! A$%! 71@'! \:"g=gh";! 5&! 'b.! H$a3K!

(P ) : x − 2y + 2z −1= 0
*!i12.!H$IJ3K!.L?3$!'b.!AjC!'b.!AjC!:S;!.R'!\!5&!.12H!ekA!5<1!'b.!H$a3K!
:l;*!>?'!.%^!7]!71@'!m!.L43!.LcA!df!(F%!A$%!'b.!H$a3K!:l;!71!nCF!.LC3K!71@'!APF!\m*!
y;!i?!:S;!.12H!ekA!5<1!:l;!343!0-3!_N3$!0t3K!_$%,3K!A-A$!.†!\!723!:l;Z!

>F!ABx!

d (I ;(P )) =
1− 2.2+ 2.(−1) −1
1
2
+ (−2)
2
+ 2
2
= 2
*!!
y;!i?!5YW!

(S ) : (x −1)
2
+ ( y − 2)
2
+ (z +1)
2
= 4
*!!!
y;!G[1!m:‡g‡g';!.$C]A!df!.%^!7]!.LC3K!71@'!APF!\m!D&!

H (
1
2
;1;
m −1
2

)
*!
V%!o!.$C]A!:l;!343!

1
2
− 2.1+ 2.
m −1
2
−1 = 0 ⇔ m −
7
2
= 0 ↔ m =
7
2
⇒ M (0;0;−
7
2
)
*!!!
!"#$%&'('%)*%+",+% /0%&'1%2%+3456%)789%+3:83%;<=%
">?@A8B6%CDEF%GFF%GCG%% )H89%IJ%83K=%L%3MN%OA83%83P8%QR%STA%3MN%U3V%%%
03A%?AW?6%X<?3@A8IOYZ8!
6!
0[R%\%]^_C%SA`=aY%Trong!mặt!phẳng!với!trục!toạ!độ!Oxy!cho!hình!chữ!nhật!ABCD!có!tâm!I.!Gọi!
M!là!trung!điểm!cạnh!CD,!H!là!hình!chiếu!vuông!góc!của!D!trên!AM!và!N!là!trung!điểm!AH.!
Tìm!toạ!độ!các!đỉnh!B,D!biết!phương!trình!đường!tròn!ngoại!tiếp!tam!giác!IMN!là!

(C ) :(x −
5

2
)
2
+ ( y −
9
2
)
2
=
5
2
!và!đỉnh!D!có!hoành!độ!nguyên!nằm!trên!đường!thẳng!

x − 2y = 0
.!
!
Đường!tròn!(C)!có!tâm

J (
5
2
;
9
2
)
,!bán!kính!

R =
10
2

.!
!!+)!Ta!có!tứ!giác!IMDN!nội!tiếp!đường!tròn:!
Chứng)minh.)
Gọi!E!là!trung!điểm!đoạn!HD,!ta!có!

NE = IM =
1
2
AD
NE / /IM
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
!nên!
tứ!giác!IMEN!là!hình!bình!hành.!
Do!đó!E!là!trực!tâm!tam!giác!MND,!và!

EM ⊥ ND ⇒ IN ⊥ ND
.!!!
Hay!góc!

IND
!
= IMD

!
= 90
0
,!do!vậy!IMDN!nội!tiếp!đường!tròn.!!
+)!Do!D!thuộc!(C)!nên!toạ!độ!D!thoả!mãn!hệ:

(x −
5
2
)
2
+ ( y −
9
2
)
2
=
5
2
y −2x = 0
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇔

x = 3, y = 6
x =
8
5
, y =
16
5
⎡
⎣
⎢
⎢
⎢
⎢
⇒ D(3;6)
.!
+)!Do!J!là!trung!điểm!của!ID!nên!I(2;3),!và!I!là!trung!điểm!của!BD!nên!B(1;0).!
!W?%@RP86!Vậy!

B(1;0),D(3;6)
.!!
! !
03b%JY!Nếu!cho!thêm!giả!thiết!điểm!

H (
63
17
;
82
17
)

!ta!tìm!được:

A(−1;2),B(1;0),C (5;4),D(3;6)
.!!
0[R%D%]C_c%SA`=aY!Tìm!số!hạng!chứa!

x
10
!trong!khai!triển!

(
nx
5
−
1
x
2
)
n
!biết!n!là!số!tự!nhiên!thoả!
mãn!

n
2
+C
n
2
= 145 (n ≥ 2)
.!!
+)!Ta!có:!

!

n
2
+
n(n −1)
2
= 145 ⇔ 3n
2
−n − 290 = 0 ⇔
n =10(t / m)
n = −
29
3
(l )
⎡
⎣
⎢
⎢
⎢
⎢
.!
+)!Vì!vậy!

(
nx
5
−
1
x

2
)
n
= (2x −
1
x
2
)
10
= C
10
k
(2x )
10−k
(−
1
x
2
)
k
k=0
10
∑
= C
10
k
.(−1)
k
.2
10−k

.x
10−3k
k=0
10
∑
.!
Chọn!

10−3k =10 ⇔ k = 0
suy!ra!số!hạng!cần!tìm!là!

C
10
0
.2
10
= 1024
.!!!!!
!
!"#$%&'('%)*%+",+% /0%&'1%2%+3456%)789%+3:83%;<=%
">?@A8B6%CDEF%GFF%GCG%% )H89%IJ%83K=%L%3MN%OA83%83P8%QR%STA%3MN%U3V%%%
03A%?AW?6%X<?3@A8IOYZ8!
7!
0[R%^C%]^_C%SA`=aY!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn!

z = min x, y,z
{ }
.!Tìm!giá!trị!nhỏ!
nhất!của!biểu!thức!

P =
x
y + z
+
y
z + x
+
3
2
.
z
x + y
−
z
z
2
+ xy + yz + zx
.!
Đặt!

a =
x
z
,b =
y
z
(a,b ≥1)
!khi!đó:!
!

P =
a
b +1
+
b
a +1
+
3
2(a + b)
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
−
1
(a +1)(b +1)
.!
+)!Ta!có:!

(a +1)(b +1) ≥ 2,∀a,b ≥1⇒ −
1
(a +1)(b +1)
≥−
1
2
,!và!!

!

a
b +1
+
b
a +1
+
3
2(a + b)
= (a + b +1)(
1
b +1
+
1
a +1
) +
3
2(a + b)
− 2
≥
4(a + b +1)
a + b + 2
+
3
2(a + b)
− 2
.!
Đặt!

t = a + b ≥ 2
,!ta!có:!!!

a
b +1
+
b
a +1
+
3
2(a + b)
≥ f (t) =
4(t +1)
t + 2
+
3
2t
− 2 ≥ f (2) =
7
4
.!
Do!đó!

P ≥ f (t )−
1
2
≥
5
4
.!!Dấu!bằng!đạt!tại!

x = y = z
.!Vậy!giá!trị!nhỏ!nhất!của!P!bằng!5/4.!!!
0dN3%G6!Sử!dụng!bất!đẳng!thức!AM!–GM!ta!có:!!

z
z
2
+ xy + yz + zx
=
z
(z + x )(z + y)
≤
1
2
z
z + x
+
z
z + y
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟

⎟
⎟
.!
Suy!ra!!

x
y + z
+
y
z + x
−
z
z
2
+ xy + yz + zx
≥
x
y + z
+
y
z + x
−
1
2
z
z + x
+
z
z + y
⎛

⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
=
1
2
2 y − z
z + x
+
2x − z
z + y
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟

⎟
.!
Sử!dụng!bất!đẳng!thức!Cauchy!–!Shcwarz!ta!có:!
!

2y − z
z + x
+
2x − z
z + y
=
(2y − z )
2
(2y − z )(z + x )
+
(2x − z)
2
(2x − z)(z + y)
≥
(2y − z + 2x − z)
2
(2y − z )(z + x ) + (2x − z)(z + y)
=
4(x + y − z)
2
4xy + z(x + y) − 2z
2
≥
4(x + y − z)
2

(x + y)
2
+ z(x + y)−2z
2
=
4(x + y − z)
x + y + 2z
.!
Vì!vậy!

P ≥
2(x + y − z )
x + y + 2z
+
3
2
.
z
x + y
.!Đặt!

t =
x + y
z
≥ 2
ta!có!

P ≥ f (t ) =
2(t −1)
t + 2

+
3
2t
.!
Xét!hàm!số!f(t)!ta!có:

f '(t) =
6
(t + 2)
2
−
3
2t
2
=
9t
2
−12t −12
t
2
(t + 2)
2
=
3(t −2)(3t + 2)
t
2
(t + 2)
2
≥ 0,∀t ≥ 2
.!!

!
!
!"#$%&'('%)*%+",+% /0%&'1%2%+3456%)789%+3:83%;<=%
">?@A8B6%CDEF%GFF%GCG%% )H89%IJ%83K=%L%3MN%OA83%83P8%QR%STA%3MN%U3V%%%
03A%?AW?6%X<?3@A8IOYZ8!
8!
!
Vì!vậy!

P ≥ f (t ) ≥ f (2) =
5
4
.!Dấu!bằng!đạt!tại!

x = y = z
.!Vậy!giá!trị!nhỏ!nhất!của!P!bằng!5/4.!
e:A%?PU%?Qf89%?g%h%%
e:A%Oi%C^6%Cho!x,y,z!là!các!số!thực!dương!thoả!mãn!

x,y, z ≤1; x + y ≥ z +1
.!Tìm!giá!trị!nhỏ!nhất!
của!biểu!thức!

P =
x
y + z
+
y
z + x
−

z
z
2
+ xy + yz + zx
.!Đ/s:!

P
min
=
1
2
.!!
e:A%Oi%CG6!Cho!x,y,z!là!các!số!thực!dương!thoả!mãn!

z = min x, y,z
{ }
.!Tìm!giá!trị!nhỏ!nhất!của!
biểu!thức!

P =
x
y + z
+
y
z + x
+
5
4
.
z

x + y
−
z
z
2
+ xy + yz + zx
.!Đ/s:!

P
min
=
9
8
.!!!!!
!
!
!