đề thi thử thpt quốc gia môn toán,đề số 25

!
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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[).\]\*)
L4^%)1%E&)$U6)_U%()`a*)@4b#c)24W&1)2d)#4^%)1%E&)1%E")>K)
e%f&)4g)>0&1)23)24"I)489)Q)!"#$%&'()*+,-) ).*.)Q)B4%)#%C#()hhhF6E#4$%&2:FG&))
Bi=)`)j.c*)>%d6kF!#$%!$&'!()!

y =
−2x−1
x +1
(1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* =>'!.%?!7@!71A'!B!.$C@D!:";!DE!$%&3$!7@!3FCG43!D-D$!7HC!.1I'!DJ3!7K3F!5&!7LM3F!.$N3F!

Δ : 3x− 4y−19 = 0
*!
Bi=).)j`c*)>%d6kF)
O; #$%!FED!O!.$%,!'P3!

tana =
1
2
*!=Q3$!F1-!.R9!DSO!01AC!.$KD!

A =
1− cosa + cos2a
sin2a − sina

*!
0; #$%!()!T$KD!U!.$%,!'P3!

(2+ i).z = 2+11i
*!=Q3$!'V7C3!DSO!U*!!!!
Bi=)7)j*c\)>%d6kF)W1,1!T$LX3F!.R>3$!

(x +1)
log
3
(x+2)+1
= (x
2
+ 2x +1)
log
9
(x+2)
2
*!!
Bi=)l)j`c*)>%d6kF!W1,1!0Y.!T$LX3F!.R>3$!

2
2
x +1
+ x ≤ 2 2+
x
(x +1)
2
*!
Bi=)\)j`c*)>%d6kF!=Q3$!.QD$!T$Z3!

I =
(x −1)
2
+ x
x
dx
1
2
∫
*!!
Bi=)-)j`c*)>%d6kF!#$%!$>3$!D$ET![*\]#^!DE!7-G!\]#^!_&!$>3$!5CV3F!D?3$!O!5&!D?3$!043![\!
5CV3F!FED!5`1!'a.!T$N3F!:\]#^;*!Ba.!T$N3F!:b;!71!cCO!\d!5CV3F!FED!5`1![#!De.!D?3$![]!.?1!f!
.$%,!'P3!

SE.SB = 2a
2
*!=Q3$!.$A!.QD$!g$)1!D$ET![*\]#^!5&!g$%,3F!D-D$!71A'!f!723!'a.!T$N3F!
:[#^;*!!!
Bi=),)j`c*)>%d6kF!=R%3F!'a.!T$N3F!.%?!7@!hiG!D$%!.O'!F1-D!\]#!DE!

AC = 2AB
5&!7j3$!#:k"lmkn;*!
=12T!.CG23!.?1!\!DSO!7LM3F!.Ro3!3F%?1!.12T!.O'!F1-D!\]#!De.!7LM3F!.$N3F!]#!.?1!71A'!p:lm";*!
=>'!.%?!7@!D-D!7j3$!\d]!012.!\!DE!$%&3$!7@!Z'!5&!T$LX3F!.R>3$!7LM3F!.$N3F!\p!
_&

x + 2y− 7= 0
*!!!!!!
Bi=)a)j`c*)>%d6kF!=R%3F!g$V3F!F1O3!5`1!$I!.RqD!.%?!7@!hiGU!D$%!$O1!71A'!\:"mlmr;!5&!]:smsmt;d!

7LM3F!.$N3F!

d :
x +1
2
=
y−1
−1
=
z
2
*!=Q3$!DV(13!FED!F1uO!7LM3F!.$N3F!\]!5&!7LM3F!.$N3F!v*!=>'!
.%?!7@!71A'!B!.R43!v!7A!D$C!51!.O'!F1-D!\B]!3$w!3$Y.*!!
Bi=)+)j*c\)>%d6kF!xO1!3FLM1!$y3!FaT!3$OC!z!.$L!51I3!.{!"r$!723!""$!(-3Fd!$|!.$)3F!3$Y.!5`1!
3$OC!32C!3FLM1!723!.RL`D!7}1!3FLM1!723!(OC!cC-!"r!T$~.!.$>!RM1!71*!=Q3$!i-D!(CY.!7A!$O1!3FLM1!
71!3F•C!3$143!'&!FaT!3$OC*!
Bi=)`*)j`c*)>%d6kF!#$%!idGdU!_&!D-D!()!.$/D!.$C@D!3€O!g$%,3F!

1;+∞
⎡
⎣
⎢
)
*!=>'!F1-!.R9!3$w!3$Y.!DSO!
01AC!.$KD!

P =
1
(x
3

+1)
2
+
1
( y
3
+1)
2
+
1
(z
3
+1)
2
−
3
2x
2
y
2
z
2
+ 2
*!
!
mmm!nLmmm)
!
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B4%)#%C#()DE#4$%&2:FG&) !
<!

)
M!oV)LpB!)qrV!)estV)/uM)uV)
!!!
Bi=)`)j.c*)>%d6kF!#$%!$&'!()!

y =
−2x−1
x +1
(1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* =>'!.%?!7@!71A'!B!.$C@D!:";!DE!$%&3$!7@!3FCG43!D-D$!7HC!.1I'!DJ3!7K3F!5&!7LM3F!.$N3F!

Δ : 3x− 4y−19 = 0
*!
"* x|D!(13$!./!F1,1*!
<* W|1!

M(m;
−2m−1
m +1
)∈ (1),m ≠ −1
*!
•;!=1I'!DJ3!7K3F!_&!

x +1= 0 ⇒ d(M;TCD)= m +1
*!!
•;!=O!DE!

d(M;Δ)=

3m − 4.
−2m−1
m +1
−19
3
2
+ (−4)
2
=
3m
2
−8m−15
5 m +1
*!
•;!=$‚%!0&1!RO!.O!DEƒ!!!
!

m +1 =
3m
2
−8m−15
5 m +1
⇔ 5(m +1)
2
= 3m
2
−8m−15
⇔
5(m +1)
2

= 3m
2
−8m−15
5(m +1)
2
= −3m
2
+ 8m +15
⎡
⎣
⎢
⎢
⎢
⇔
m =1(t / m)
m = −
5
4
(l)
m =
−9± 41
2
(l)
⎡
⎣
⎢
⎢
⎢
⎢
⎢

⎢
⎢
⎢
⎢
⎢
⎢
⇒ M(1;−
3
2
)
*!
„JG!71A'!D…3!.>'!

M(1;−
3
2
)
*!!
Bi=).)j`c*)>%d6kF)
O; #$%!FED!O!.$%,!'P3!

tana =
1
2
*!=Q3$!F1-!.R9!DSO!01AC!.$KD!

A =
1− cosa + cos2a
sin2a − sina
*!

0; #$%!()!T$KD!U!.$%,!'P3!

(2+ i).z = 2+11i
*!=Q3$!'V7C3!DSO!U*!!!!
O; =O!DEƒ!

A =
1− cosa + 2cos
2
a −1
sina(2cosa −1)
=
cosa(2cosa −1)
sina(2cosa −1)
= cota =
1
tana
= 2
*!!
0; =O!DEƒ!

z =
2+11i
2+ i
=
(2+11i)(2− i)
5
=
15+ 20i
5

= 3+ 4i ⇒ z = 3− 4i
*!
„>!5JG!

z = 3
2
+ (−4)
2
= 5
*!!!
Bi=)7)j*c\)>%d6kF)W1,1!T$LX3F!.R>3$!

(x +1)
log
3
(x+2)+1
= (x
2
+ 2x +1)
log
9
(x+2)
2
*!!
!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&) !
s!
†1HC!g1I3ƒ!

x >−2
*!
b$LX3F!.R>3$!.LX3F!7LX3F!5`1ƒ!!
!

(x +1)
log
3
(x+2)+1
= (x +1)
2log
9
(x+2)
2
⇔
x +1= 1
0< x +1≠ 1
log
3
(x + 2)+1= 2log
9
(x + 2)
2
⎧
⎨
⎪
⎪
⎪
⎩
⎪

⎪
⎪
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⇔
x = 0
x = 1
⎡
⎣
⎢
⎢
⎢
*!
„JG!T$LX3F!.R>3$!DE!$O1!3F$1I'!

x = 0;x = 1
*!!
Bi=)l)j`c*)>%d6kF!W1,1!0Y.!T$LX3F!.R>3$!

2
2
x +1
+ x ≤ 2 2+
x
(x +1)

2
*!
†1HC!g1I3ƒ!

x ≥ 0
*!
]Y.!T$LX3F!.R>3$!.LX3F!7LX3F!5`1ƒ!

2 2(x +1)+(x +1) x ≤ 2 2x
2
+ 5x + 2 ⇔ 4(2x
2
+ 5x + 2)≥(2 2(x +1) + (x +1) x )
2
⇔ 4(x +1) 2(x
2
+ x) + x
3
−6x
2
−11x ≤ 0 ⇔ x(x
2
−2x +1)+ 4(x +1) 2(x
2
+ x) − 4x(x + 3)≤ 0
⇔ x(x−1)
2
+ 4 x (x +1) 2(x +1) −(x + 3) x
⎡
⎣

⎢
⎢
⎤
⎦
⎥
⎥
≤ 0 ⇔ x(x−1)
2
+ 4 x .
2(x +1)
3
− x(x + 3)
2
(x +1) 2(x +1) + (x + 3) x
≤ 0
⇔ x(x−1)
2
+
4 x (x −1)
2
(x + 2)
(x +1) 2(x +1) + (x + 3) x
≤ 0 ⇔ x (x −1)
2
x +
4(x + 2)
(x +1) 2(x +1) + (x + 3) x
⎡
⎣
⎢

⎢
⎢
⎤
⎦
⎥
⎥
⎥
≤ 0
⇔ x (x−1)
2
= 0 ⇔
x = 0
x = 1
⎡
⎣
⎢
⎢
⎢
(do x +
4(x + 2)
(x +1) 2(x +1) + (x + 3) x
> 0,∀x ≥ 0)
*!
Kết$luận:!„JG!.JT!3F$1I'!DSO!0Y.!T$LX3F!.R>3$!_&!

S = 0;1
{ }
*!
Cách$2:!]Y.!T$LX3F!.R>3$!.LX3F!7LX3F!5`1ƒ!

8
x +1
+ x + 4
2x
x +1
≤ 4(2+
x
(x +1)
2
) ⇔ 4
2x
x +1
≤
x(−x
2
+ 6x +11)
(x +1)
2
*!
•;!‡2C!

x = 0
d!0Y.!T$LX3F!.R>3$!_CV3!7~3F*!
•;!!‡2C!

x > 0
d!0Y.!T$LX3F!.R>3$!.LX3F!7LX3F!5`1ƒ!

−x
2

+ 6x +11
(x +1)
2
≥ 4
2
x(x +1)
(1) ⇔
−x
2
+ 6x +11> 0
(−x
2
+ 6x +11)
2
(x +1)
4
≥
32
x(x +1)
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪

⇔
−x
2
+ 6x +11> 0
x(−x
2
+ 6x +11)
2
−32(x +1)
3
≥ 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇔
−x
2
+ 6x +11> 0
x
5
−12x
4
−18x
3
+ 36x
2

+ 25x −32≥ 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇔
−x
2
+ 6x +11> 0
(x −1)
2
(x
3
−10x
2
−39x−32)≥ 0
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇔
−x
2
+ 6x +11> 0

(x −1)
2
. −x(−x
2
+ 6x +11)−4x
2
−28x−32
⎡
⎣
⎢
⎤
⎦
⎥
≥ 0
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇔ x = 1(do − x(−x
2
+ 6x +11)−4x
2
−28x−32< 0,∀x ∈ (0;3+ 2 5))
*!
Kết$luận:!„JG!.JT!3F$1I'!DSO!0Y.!T$LX3F!.R>3$!_&!

S = 0;1
{ }
*!
!
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B4%)#%C#()DE#4$%&2:FG&) !
ˆ!
qv&4)$=;&F!]Y.!T$LX3F!.R>3$!5`1!_M1!F1,1!.R43!.O!D$j!0123!7‰1!.LX3F!7LX3F!0Š3F!T$‹T!0>3$!
T$LX3F!$O1!52!7LO!5H!0Y.!T$LX3F!.R>3$!DX!0,3!

B ≥ A
*!‡F%&1!.O!.O!DE!.$A!.$/D!$1I3!D-D!D-D$!
g$-D!(OCƒ!
Cách$2:![€!vq3F!0Y.!7N3F!.$KD!\B!ŒWB!.O!DE!
!

4
2
x(x +1)
=
8
2x(x +1)
≥
8
2x + x +1
2
=
16
3x +1

*!
‡43!.{!:";!(CG!ROƒ!
!

−x
2
+ 6x +11
(x +1)
2
≥
16
3x +1
⇔ (3x +1)(−x
2
+ 6x +11)−16(x +1)
2
≥ 0
⇔ −(x −1)
2
(x +
5
3
)≥ 0 ⇔ x = 1(do x > 0)
*!
=O!DE!g2.!cC,!.LX3F!./!.R43*!
Cách$3:!=O!DE!:";!.LX3F!7LX3F!5`1ƒ!

(
8
2x(x +1)

−
16
3x +1
)+
3x +17
3x +1
−
4(2x + 3)
(x +1)
2
≤ 0
⇔
8(3x +1−2 2x(x +1))
(3x +1) 2x(x +1)
+
(3x +17)(x +1)
2
− 4(2x +3)(3x +1)
(3x +1)(x +1)
2
≤ 0
⇔
8( 2x − x +1)
2
(3x +1) 2x(x +1)
+
(x −1)
2
(3x + 5)
(3x +1)(x +1)

2
≤ 0 ⇔
2x = x +1
x −1= 0
⎡
⎣
⎢
⎢
⎢
⇔ x = 1
*!
Bi=)\)j`c*)>%d6kF!=Q3$!.QD$!T$Z3!

I =
(x −1)
2
+ x
x
dx
1
2
∫
*!!
=O!DEƒ!

I =
x
2
−2x +1+ x
x

dx
1
2
∫
= (x −2+
1
x
+
1
x
)dx
1
2
∫
= (
x
2
2
−2x + ln x + 2 x )
2
1
= −
5
2
+ ln2+ 2 2
*!
Bi=)-)j`c*)>%d6kF!#$%!$>3$!D$ET![*\]#^!DE!7-G!\]#^!_&!$>3$!5CV3F!D?3$!O!5&!D?3$!043![\!
5CV3F!FED!5`1!'a.!T$N3F!:\]#^;*!Ba.!T$N3F!:b;!71!cCO!\d!5CV3F!FED!5`1![#!De.!D?3$![]!.?1!f!
.$%,!'P3!

SE.SB = 2a
2
*!=Q3$!.$A!.QD$!g$)1!D$ET![*\]#^!5&!g$%,3F!D-D$!71A'!f!723!'a.!T$N3F!
:[#^;*!!!
=O!DEƒ!

BC ⊥ AB
BC ⊥ SA
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ BC ⊥ (SAB)⇒ BC ⊥ AE
*!
[CG!RO!

AE ⊥ BC
AE ⊥ SC(gt)
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ AE ⊥ (SBC)⇒ AE ⊥ SB
*!!!

„JG!\f!_&!7LM3F!DO%!.R%3F!.O'!F1-D![\]d!5>!5JG!

SE.SB = SA
2
= 2a
2
⇒ SA = a 2
*!
„&!

V
S.ABCD
=
1
3
SA.S
ABCD
=
1
3
.a 2.a
2
=
a
3
2
3
:75 ;*!
•;!=Q3$!v:fm:[#^;;*!
!

!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&) !
"!
#$!%&!

SB = SA
2
+ AB
2
= 2a
2
+ a
2
= a 3 ⇒ SE =
SA
2
SB
=
2a
2
a 3
=
2a 3
3
'!
()!*+,!

SE
SB
=

2a 3 / 3
a 3
=
2
3
⇒ d(E;(SCD))=
2
3
d(B;(SCD)) (1)
'!
/!%&!012234567!898!

d(B;(SCD)) = d(A;(SCD)) (2)
'!
:;!0<!*=>8?!?&%!*@/!46!A./!<!AB)!

AH ⊥ (SCD) ⇒ AH = d(A;(SCD)) (3)
'!
#C!3D7E3F7!*G!3H7!I=,!J$K!

d(E;(SCD))=
2
3
AH
'!
#$L!?/M%!*=>8?!406!%&!

1
AH
2

=
1
SA
2
+
1
AD
2
=
1
2a
2
+
1
a
2
=
3
2a
2
⇒ AH =
a 6
3
'!
(+,!

d(E;(SCD))=
2a 6
9
'!!!!!!!!!

HI%)#;@)#<J&1)#KL!5BN!B)8B!%B&O!4'0156!%&!PM,!0156!QG!B)8B!*=>8?!%.8B!$!*G!%.8B!R98!40!
*=>8?!?&%!*@/!LSA!OBT8?!301567'!USA!OBT8?!3V7!P/!W=$!0E!*=>8?!?&%!*@/!45!%XA!%M%!%.8B!
41E45E46!QY8!QZ[A!A./!\E]E<!ABN^!L_8!

S
AEFH
=
a
2
2
3
'!#`8B!ABa!A`%B!bBc/!%B&O!4'0156!*G!bBN^8?!
%M%B!P/aL!\!Pd8!LSA!OBT8?!34567'!!e2IK!!

V
S.ABCD
=
a
3
2
3
;d(E;(SCD))=
2a 6
9
'!
BM=),)NOP*)>%Q6RF!#JN8?!LSA!OBT8?!AN.!Pf!gh,!%BN!A$L!?/M%!015!%&!

AC = 2AB
*G!Pi8B!53jD"kjl7'!
#/dO!A=,d8!A./!0!%m$!PZn8?!AJo8!8?N./!A/dO!A$L!?/M%!015!%XA!PZn8?!ABT8?!15!A./!P/aL!p3"kD7'!

#)L!AN.!Pf!%M%!Pi8B!0E1!R/dA!0!%&!BNG8B!Pf!qL!*G!OBZr8?!AJ)8B!PZn8?!ABT8?!0p!
QG

x + 2y− 7= 0
'!!!!!!
stA!B$/!A$L!?/M%!10p!*G!05p!%&K

BAI
!
= C
"
,I
#
(chung)
!898!B$/!A$L!?/M%!10p!*G!05p!Pu8?!v.8?'!
()!*+,!

AB
AC
=
IB
IA
⇒
AB
2
AC
2
=
IB
2

IA
2
(1)
'!
USA!bBM%E!vN!p0!QG!A/dO!A=,d8!898!

IA
2
= IB.IC (2)
'!
#C!3D7E3F7!A$!%&K!

IB
IC
=
AB
2
AC
2
=
1
4
⇒ IB
!"!
=
1
4
IC
! "!
=

1
4
(−20;−10) = (−5;−
5
2
)⇒ B(0;−
3
2
)
'!
w7!xy/!

A(7− 2a;a)∈ AI
E!A$!%&K!!

AC
2
= 4AB
2
⇔ (22−2a)
2
+ (a + 9)
2
= 4 (7− 2a)
2
+ (a +
3
2
)
2

⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⇔ 15(a
2
− 2a − 24)= 0 ⇔
a = 6(l)
a = −4(t / m)
⎡
⎣
⎢
⎢
⎢
⇒ A(15;−4)
'!
SC#)$=;&(!(+,!03D"kjz7E!13{kjH2F7'!!!!!
HT&4)$=;&(!#$!%&!ABa!A)L!8B$8B!P/aL!0!%|8?!AC!B$/!A$L!?/M%!10p!*G!05p!Pu8?!v.8?!8BZ!I$=K!

AB
AC
=
IB
IA

=
IA
IC
⇒ IC = 2IA ⇒ A
'!
!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&) !
}!
BM=)U)NOP*)>%Q6RF!#JN8?!bB>8?!?/$8!*@/!B~!AJ•%!AN.!Pf!gh,€!%BN!B$/!P/aL!03Dk"k{7!*G!13HkHk}7E!
PZn8?!ABT8?!

d :
x +1
2
=
y−1
−1
=
z
2
'!#`8B!%>I/8!?&%!?/•$!PZn8?!ABT8?!01!*G!PZn8?!ABT8?!v'!#)L!
AN.!Pf!P/aL!U!AJ98!v!Pa!%B=!*/!A$L!?/M%!0U1!8B‚!8BƒA'!!
#$!%&K!

AB
! "!
= (2;−2;6) / /(1;−1;3)
E!PZn8?!ABT8?!v!%&!*t%!Ar!%Bi!OBZr8?!

u
!
= (2;−1;2)
'!
()!*+,!

cos(AB;d)
!
=
1.2+ (−1).(−1)+ 3.2
1
2
+ (−1)
2
+ 3
2
. 2
2
+ (−1)
2
+ 2
2
=
3
11
'!
w7!xy/!

M(−1+ 2t;1− t;2t)∈ d ⇒ MA = 9t
2

+ 20,MB = 9t
2
− 36t + 56,AB = 2 11
'!
6N!01!bB>8?!P„/!898!%B=!*/!A$L!?/M%!0U1!8B‚!8BƒA!bB/!U0wU1!8B‚!8BƒA'!
#$!%&K!

MA + MB = 9t
2
+ 20 + 9t
2
− 36t + 56 = (3t)
2
+ (2 5)
2
+ (6− 3t)
2
+ (2 5)
2
'!
stA!B$/!*t%!Ar!

a
!
= (3t;2 5),b
!
= (6− 3t;2 5)
E!I…!v•8?!RƒA!PT8?!AB†%K!

a

!
+ b
!
≥ a
!
+ b
!
!E!A$!%&!!
!

MA + MB ≥ (3t + 6−3t)
2
+ (2 5+ 2 5)
2
= 2 29
'!
6ƒ=!R‡8?!h^,!J$!bB/!*G!%Bi!bB/!

a
!
↑↑ b
!
⇔
3t
6−3t
=
2 5
2 5
⇔ t = 1⇒ M(1;0;2)
'!

(+,!P/aL!%Y8!A)L!U3Dk{kF7'!!
BM=)+)N*PV)>%Q6RF!<$/!8?Zn/!Bˆ8!?SO!8B$=!‰!ABZ!*/~8!AC!ŠB!Pd8!lB!IM8?E!By!ABc8?!8BƒA!*@/!8B$=!
8d=!8?Zn/!Pd8!AJZ@%!P[/!8?Zn/!Pd8!I$=!W=M!D{!OB‹A!AB)!Jn/!P/'!#`8B!hM%!I=ƒA!Pa!B$/!8?Zn/!P/!
8?Œ=!8B/98!LG!?SO!8B$='!
!
w7!x/^!I…!h!3OB‹A7!QG!ABn/!?/$8!LG!8?Zn/!AB†!8BƒA!
%Bn!8?Zn/!AB†!B$/!‰!ABZ!*/~8'!
xy/!,!3OB‹A7!QG!ABn/!?/$8!LG!8?Zn/!AB†!B$/!%Bn!
8?Zn/!AB†!8BƒA!‰!ABZ!*/~8'!
xy/!0!QG!R/d8!%c!B$/!8?Zn/!?SO!8B$='!
e/•=!b/~8K!

0 ≤ x,y ≤ 60
E!898!bB>8?!?/$8!LŒ=!QG!
B)8B!*=>8?!%&!v/~8!A`%B!R‡8?!

S = 60
2
= 3600
3Pr8!
*Ž7'!
w7!ea!B$/!8?Zn/!?SO!8B$=!A$!OB^/!%&K!

x− y ≤10 ⇔ −x +10≤ y ≤ x +10 (*)
'!
5M%!P/aL!U3hk,7!ABN^!L_8!P/•=!b/~8!3•7!QG!OBY8!
?.%B!Iy%!AJ98!B)8B!*•!3R98!vZ@/7!*G!QG!Ic!OBY8!A…!
%m$!R/d8!%c!0'!VBY8!?.%B!Iy%!8G,!%&!v/~8!A`%B!

S '= S −(

1
2
50
2
+
1
2
50
2
)= 3600−50
2
= 1100
3Pr8!*Ž7'!
()!*+,!

P(A)=
S '
S
=
1100
3600
=
11
36
'!
! !
!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&) !
‘!

BM=)O*)NOP*)>%Q6RF!5BN!hE,E€!QG!%M%!Ic!AB’%!vZr8?!AB=f%!

1;+∞
⎡
⎣
⎢
)
'!#)L!?/M!AJŽ!8B‚!8BƒA!%m$!R/a=!
AB†%!

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

2x
2
y
2
z
2
+ 2
'!
W4;&)XY#(!(@/!$ER!QG!B$/!Ic!AB’%!vZr8?!A$!%&K!

1
(a +1)
2
+
1
(b +1)
2
≥
1
ab +1
'!
Chứng$minh.!4…!v•8?!RƒA!PT8?!AB†%!5$=%B,!“4%B”$JA€!A$!%&K!
!

(ab +1)(
a
b
+1) ≥ (a +1)
2
⇒

1
(a +1)
2
≥
b
(a + b)(ab +1)
,
(ab +1)(
b
a
+1) ≥ (b +1)
2
⇒
1
(b +1)
2
≥
a
(a + b)(ab +1)
'!
5f8?!AB•N!*d!B$/!RƒA!PT8?!AB†%!AJ98!A$!PZ[%K!
!

1
(a +1)
2
+
1
(b +1)
2

≥
a
(a + b)(ab +1)
+
b
(a + b)(ab +1)
=
1
ab +1
'!
6ƒ=!R‡8?!h^,!J$!bB/!*G!%Bi!bB/!

a = b = 1
'!
–O!v•8?!A$!%&K!!!!

1
(x
3
+1)
2
+
1
( y
3
+1)
2
≥
1
x

3
y
3
+1
,
1
( y
3
+1)
2
+
1
(z
3
+1)
2
≥
1
y
3
z
3
+1
,
1
z
3
+1
+
1

x
3
+1
≥
1
z
3
x
3
+1
'!
5f8?!AB•N!*d!R$!RƒA!PT8?!AB†%!AJ98!A$!PZ[%K!
!

1
(x
3
+1)
2
∑
≥
1
2
1
x
3
y
3
+1
∑

≥
3
2(x
2
y
2
z
2
+1)
'!
B4Z)3(!(@/!Ly/!

a,b,c ≥1
A$!%&K!
!

1
a
3
+1
+
1
b
3
+1
+
1
c
3
+1

≥
3
abc +1
3•7'!
Chứng$minh.!#$!%&!RƒA!PT8?!AB†%!W=•8!AB=f%K!!
!

1
1+ m
2
+
1
1+ n
2
≥
2
1+ mn
,∀ mn ≥1
'!
–O!v•8?!A$!%&K!!
!

1
1+ a
3
+
1
1+ b
3
≥

2
1+ a
3
b
3
,
1
1+ c
3
+
1
1+ abc
≥
2
1+ abc
4
,
2
1
1+ a
3
b
3
+
1
abc
4
⎛
⎝
⎜

⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
≥
4
1+ a
3
b
3
. abc
4
=
4
1+ abc
'!
5f8?!AB•N!*d!R$!RƒA!PT8?!AB†%!AJ98!A$!%&!3•7!PZ[%!%B†8?!L/8B'!
—=$,!Q./!RG/!ANM8E!MO!v•8?!A$!%&K!

1
x
3
y
3

+1
∑
≥
3
x
2
y
2
z
2
+1
'!!!!
!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&) !
Š!
4=,!J$K

P ≥
3
2(x
2
y
2
z
2
+1)
−
3
2(x

2
y
2
z
2
+1)
=
3
2
1
x
2
y
2
z
2
+1
−
1
2
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟

⎟
⎟
⎟
⎟
⎟
2
−
3
4
≥−
3
4
!'!
6ƒ=!R‡8?!h^,!J$!bB/!*G!%Bi!bB/!

x = y = z =1
'!!
SC#)$=;&(!(+,!?/M!AJŽ!8B‚!8BƒA!%m$!V!R‡8?!

−
3
4
P.A!A./!

x = y = z =1
'!!!
Cách$2:!–O!v•8?!RƒA!PT8?!AB†%K!

a
2

+ b
2
+ c
2
≥
1
3
(a + b+ c)
2
'!#$!%&K!

1
(x
3
+1)
2
+
1
( y
3
+1)
2
+
1
(z
3
+1)
2
≥
1

3
1
x
3
+1
+
1
y
3
+1
+
1
z
3
+1
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
2
'!
–O!v•8?!RƒA!PT8?!AB†%!3•7!A$!%&K!

1
x
3
+1
+
1
y
3
+1
+
1
z
3
+1
≥
3
1+ xyz
'!!
6N!P&K!

P ≥
3
(1+ xyz)
2
−
3
2x
2
y

2
z
2
+ 2
'!!
USA!bBM%K!

2x
2
y
2
z
2
+ 2 ≥1+ xyz
E!vN!P&K!

P ≥
3
(1+ xyz)
2
−
3
1+ xyz
= 3
1
1+ xyz
−
1
2
⎛

⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
2
−
3
4
≥−
3
4
'!
HT&4)$=;&(!5B‹!˜!AJN8?!Qn/!?/^/!AJ98!A$!I…!v•8?!%M%!RƒA!PT8?!AB†%!B$,!v™8?!I$=K!
w7!5BN!8!Ic!AB’%!vZr8?!ABN^!L_8!

x
i
≥1,i = 1,n
A$!%&K!
!

1

x
1
+1
+
1
x
2
+1
+ +
1
x
n
+1
≥
n
x
1
x
2
x
n
n
+1
'!
w7!(@/!$ER!QG!B$/!Ic!AB’%!vZr8?!A$!%&K!
!

1
(a +1)
2

+
1
(b +1)
2
≥
1
ab +1
'!
#„8?!W=MA!Br8!A$!%&K!
!

1
(a + k)
2
+
1
(b + k)
2
≥
1
ab + k
2
,k > 0
'!
Chứng$minh.$
4…!v•8?!RƒA!PT8?!AB†%!5$=%B,!“4%B”$JA€!A$!%&K!
!

ab + k
2

( )
(
a
b
+1) ≥ (a +k)
2
⇒
1
(a + k)
2
≥
b
(a + b)(ab + k
2
)
,
(ab +k
2
)(
b
a
+1) ≥ (b + k)
2
⇒
1
(b +k)
2
≥
a
(a + b)(ab + k

2
)
'!
5f8?!AB•N!*d!B$/!RƒA!PT8?!AB†%!AJ98!A$!PZ[%K!
!

1
(a + k)
2
+
1
(b + k)
2
≥
b
(a + b)(ab + k
2
)
+
a
(a + b)(ab + k
2
)
=
1
ab + k
2
'!
6ƒ=!R‡8?!h^,!J$!bB/!*G!%Bi!bB/!

a = b = k
'!
!
!"#$%&'()*+,-) ).*.)) /0&1)23)&456)7)489):%&4)&4;&)<=)>?%)489)@4A)))
B4%)#%C#()DE#4$%&2:FG&) !
l!
ea!%B/!A/dA!Br8!*•!%M%!RG/!ANM8!MO!v•8?!B$/!RƒA!PT8?!AB†%!OB•!8G,!1.8!Py%!AB$L!bB^N!5=c8!
š:›!AB=+A!?/^/!1ƒA!PT8?!AB†%!1G/!ANM8!U/8!“!U$hœ!%m$!%™8?!AM%!?/^'!
HI%)#;@)#<J&1)#K)L)
HI%):[)*OF)5BN!hE,E€!QG!%M%!Ic!AB’%!vZr8?!ABN^!L_8!

x, y,z ≥1
'!#)L!?/M!AJŽ!8B‚!8BƒA!%m$!R/a=!AB†%!

1
(x
3
+1)
2
+
1
( y
3
+1)
2
+
1
(z
3
+1)

2
≥
3
2(x
2
y
2
z
2
+1)
'!
HI%):[)*.F!5BN!hE,E€!QG!%M%!Ic!AB’%!vZr8?!ABN^!L_8!

x, y,z ≥ 2
'!#)L!?/M!AJŽ!8B‚!8BƒA!%m$!R/a=!AB†%!

P =
1
(x
3
+ 8)
2
+
1
( y
3
+ 8)
2
+
1

(z
3
+ 8)
2
−
3
64 2(x
2
y
2
z
2
+ 64)
'!
!
!!
!!
!
!!
!!
!
!!
!
!!!
!
!!!!!
!
!
!!!!!
!!

!!