Đề thi thử và cách giải môn toán thầy đặng thành nam (6)

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%)()*^\*.\.*_])
L4`%)1%E&)$U6)aU%()_^*)@4b#c)24W&1)2d)#4`%)1%E&)1%E")>K)
e%f&)4g)>0&1)23)24"I)489)Q)!"#$%&'()*+,-) ).*.)
Bh=)_)i.c*)>%d6jF)#$%!$&'!()!

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

m =1
*!
=* #$%!71>'!

I 0;−
8
5
⎛
⎝
⎜
⎜
⎜

⎜
⎞
⎠
⎟
⎟
⎟
⎟
*!?@'!'!7>!:";!AB!C!71>'!A/A!.D9!EFGF#!5&!

IA = IB = IC
*!!
Bh=).)i_c*)>%d6jF!!
"* H1,1!I$JK3L!.D@3$!

tan x.cot x +
π
4
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
= 1−tan x

*!
=* H1,1!I$JK3L!.D@3$!

6
2 x−x
2
+ 2 = 2
2 x−x
2
+ 2.3
2 x−x
2
*!!
Bh=)7)i_c*)>%d6jF!?M3$!.MA$!I$N3!

I =
x + x −2
x −2
dx
3
6
∫
*!
Bh=)k)i_c*)>%d6jF!!
O; ?@'!API!()!.$/A!:QRS;!.$%,!'T3!

(x − yi −1)
2
= −1
*!!

0; U2I!3LVW!3$143!X!$YA!(13$!ZO'!5&!X!$YA!(13$!Z[!.$&3$!'\.!$&3L!]YA*!?M3$!Q-A!(W^.!7>!
$YA!(13$!ZO'!5&!Z[!7_3L!Q`3!a6!3$OW*!
Bh=)])i_c*)>%d6jF)#$%!$@3$!A$BI!b*EG#c!AB!

AD = CD = a,AB = 2a, BAD
!
= ADC
!
= 90
0
*!#d3$!
043!

SA = 3a
5&!5We3L!LBA!5<1!'P.!I$f3L!:EG#c;*!HY1!g!h&!L1O%!71>'!AiO!E#!5&!Gc*!?M3$!.$>!
.MA$!a$)1!A$BI!b*EG#!5&!a$%,3L!A-A$!.j!g!723!'P.!I$f3L!:b#c;*!!
Bh=)-)i_c*)>%d6jF)?D%3L!a$e3L!L1O3!5<1!.DkA!.%d!7\!lQSm!A$%!$O1!71>'!E:"R"R=;!5&!G:"Rn"R";F!
7Jo3L!.$f3L!

d :
x −1
1
=
y −1
1
=
z +1
1
*!p12.!I$JK3L!.D@3$!'P.!I$f3L!:q;!A$_O!EG!5&!(%3L!(%3L!
5<1!]*!?M3$!a$%,3L!A-A$!.j!]!723!'P.!I$f3L!:q;*!)

Bh=),)i_c*)>%d6jF!?D%3L!'P.!I$f3L!5<1!.DkA!.%d!7\!lQS!A$%!7Jo3L!.$f3L!

d : 6x + 8 y +11 = 0
*!p12.!
I$JK3L!.D@3$!7Jo3L!.Dr3!:?;!AB!.N'!g!0-3!aM3$!0s3L!

2
!5&!At.!lQF!lS Fu Z !hv3!hJw.!.$`%!A-A!7% d3!
.$f3L!EGF#cFuZ!.$%,!'T3!

S
IMN
= 1;AB = CD (x
I
> 0)
*!
Bh=)^)i_c*)>%d6jF!H1,1!$x!I$JK3L!.D@3$!

x(x + (x − y)(x − 2y)) = y( y + 3)
2
12x + y +7( y(x − y) + y(x − 2y)) = 8
( y + 3)
2
y
⎧
⎨
⎪
⎪
⎪
⎪

⎩
⎪
⎪
⎪
⎪
*!
Bh=)+)i_c*)>%d6jF!#$%!QFSFm!h&!A-A!()!.$/A!.$%,!'T3!

x
2
+ y
2
+ z
2
= 2
*!?@'!L1-!.D9!3$y!3$^.!AiO!
01>W!.$_A!

P =
1
x − y + 2
+
1
y − z + 2
+
1
z − x + 2
+
1
xy + yz + zx + 2

*!
lll!mLlll)


!
!
=!
PHÂN TÍCH BÌNH LUẬN VÀ ĐÁP ÁN CHI TIẾT

Bh=)_)i.c*)>%d6jF)#$%!$&'!()!

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

m =1
*!
=* #$%!71>'!

I 0;−
8
5
⎛
⎝
⎜
⎜

⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
*!?@'!'!7>!:";!AB!C!71>'!A/A!.D9!EFGF#!5&!

IA = IB = IC
*!!
"* zYA!(13$!./!h&'*!
=* ?O!AB{!

y ' = 4x
3
− 4mx; y' = 0 ⇔
x = 0
x
2
= m
⎡
⎣
⎢
⎢
*!
|>!:";!AB!0O!71>'!A/A!.D9!a$1!S}!AB!0O!3L$1x'!I$N3!01x.!!

⇔ m > 0

*!
+$1!7B!.%d!7\!0O!71>'!A/A!.D9!h&!

A(0;2m −1),B (− m ;−(m −1)
2
),C ( m ;−(m −1)
2
)
*!
c%!E!.$W\A!lSR!GF#!7)1!Q_3L!~WO!lS!343!.O'!L1-A!EG#!AN3!.d1!E!5&!AB!.DkA!7)1!Q_3L!lS*!p@!g!
.$W\A!lS!343!

IB = IC
*!p•S!.$`%!S4W!AvW!0&1!.%-3!.O!A$€!Av3{!!

IA = IB ⇔ (−
8
5
− 2m +1)
2
= m + (−
8
5
+ (m −1)
2
)
2
⇔ 5m
4
+ 20m

3
− 6m
2
−19m = 0
⇔ m(m −1)(5m
2
+ 25m +19) = 0 ⇔ m = 1(do m > 0)
*!
p•S!L1-!.D9!Av3!.@'!h&!

m =1
*!!
Bh=).ikc*)>%d6j!H1,1!A-A!I$JK3L!.D@3$!!
O;

tan x.cot x +
π
4
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟

= 1−tan x
R!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!0;!

6
2 x−x
2
+ 2 = 2
2 x−x
2
+ 2.3
2 x−x
2
*!!
O; |1•W!a1x3{!

cos x ≠ 0
sin x +
π
4
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟

⎟
≠ 0
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
*!
q$JK3L!.D@3$!.JK3L!7JK3L!5<1{!

sin x
cos x
.
cos x +
π
4
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠

⎟
⎟
⎟
⎟
sin x +
π
4
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
=1−
sin x
cos x
⇔
sin x(cosx −sin x )
cos x (sin x + cos x )
=
cos x −sin x
cos x
⇔ cos x −sin x
( )

sin x
cos x (sin x + cos x )
−
1
cos x
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
= 0
⇔ −
cos x −sin x
sin x + cos x
= 0 ⇔ cosx = sin x ⇔ tan x =1 ⇔ x =
π
4
+ kπ
*!
p•S!3L$1x'!AiO!I$JK3L!.D@3$!h&!

x =
π
4
+ kπ,k ∈ !
*!!!
0; q$JK3L!.D@3$!.JK3L!7JK3L!5<1{!

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

2
=1
⎡
⎣
⎢
⎢
⎢
⇔
x = 0
x = 2
x =1
⎡
⎣
⎢
⎢
⎢
⎢
⎢
*!
p•S!I$JK3L!.D@3$!AB!0O!3L$1x'!

x = 0;x = 1;x = 2
*!
Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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B4%)#%C#()DE#4$%&2:FG&)!
C!
Bh=)7)i_c*)>%d6jF!?M3$!.MA$!I$N3!

I =

x + x −2
x −2
dx
3
6
∫
*!
?O!AB{!

I =
x −2 + 2 + x −2
x −2
dx
3
6
∫
= 1+
2
x −2
+
1
x −2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠

⎟
⎟
⎟
⎟
dx
3
6
∫
= x + 2ln x −2 + 2 x −2
( )
6
3
= 5+ 4 ln 2
*!!
Bh=)k)i_c*)>%d6jF!!
O; ?@'!API!()!.$/A!:QRS;!.$%,!'T3!

(x − yi −1)
2
= −1
*!!
0; U2I!3LVW!3$143!X!$YA!(13$!ZO'!5&!X!$YA!(13$!Z[!.$&3$!'\.!$&3L!]YA*!?M3$!Q-A!(W^.!7>!
$YA!(13$!ZO'!5&!Z[!7_3L!Q`3!a6!3$OW*!
O; ?O!AB{!

(x − yi −1)
2
= −1= i
2
⇔

x − yi −1= −i
x − yi −1= i
⎡
⎣
⎢
⎢
⇔
x −1= 0
−y = −1
⎧
⎨
⎪
⎪
⎩
⎪
⎪
x −1= 0
−y =1
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⎡
⎣
⎢
⎢
⎢

⎢
⎢
⎢
⎢
⇔
x = 1, y = 1
x = 1, y = −1
⎡
⎣
⎢
⎢
*!
p•S!

(x; y) = (1;1);(1;−1)
*!!
0; U2I!.W‚!ƒ!„!$YA!(13$!AB!„…!#-A$*!
H1,!(†!$&3L!]YA!7JwA!7-3$!()!.j!"!723!„R!()!A-A$!Q2I!.$%,!'T3!32W!$YA!(13$!ZO'!7_3L!59!.DM!
7-3$!()!A$‡3!5&!$YA!(13$!3[!7_3L!59!.DM!7-3$!()!hˆ!$%PA!3LJwA!hd1*!
‰;!Z2W!X!$YA!(13$!ZO'!7_3L!59!.DM!7-3$!()!A$‡3!AB!X…!A-A$!Q2IR!(OW!7B!Q2I!X!Z[!5&%!59!.DM!
7-3$!()!hˆ!AB!X…!#-A$*!p•S!AB!X…*X…!A-A$*!
‰;!Z2W!X!$YA!(13$!Z[!7_3L!59!.DM!7-3$!()!A$‡3!AB!X…!A-A$!Q2IR!(OW!7B!Q2I!X!ZO'!5&%!59!.DM!
7-3$!()!hˆ!AB!X…!#-A$*!p•S!AB!X…*X…!A-A$*!
p•S!.^.!A,!AB!=*X…*X…!A-A$!Q2I!.$%,!'T3*!
p@!5•S!Q-A!(W^.!Av3!.M3$!h&!

P =
2.4!.4!
8!
=

1
35
*!
B4b)3F!?Š3L!~W !Q2I!Q`3!a6!3!$YA!(13$!ZO'!5&!3!$YA!(13$!Z[!AB!

2.(n!)
2
A-A$*!!!
Bh=)])i_c*)>%d6jF)#$%!$@3$!A$BI!b*EG#c!AB!

AD = CD = a,AB = 2a, BAD
!
= ADC
!
= 90
0
*!#d3$!
043!

SA = 3a
5&!5We3L!LBA!5<1!'P.!I$f3L!:EG#c;*!HY1!g!h&!L1O%!71>'!AiO!E#!5&!Gc*!?M3$!.$>!
.MA$!a$)1!A$BI!b*EG#!5&!a$%,3L!A-A$!.j!g!723!'P.!I$f3L!:b#c;*!!
!
‰;!?O!AB!

S
ABC
=
1
2

AB.d(C;AB ) =
1
2
AB.AD =
1
2
.2a.a = a
2
*!
p@!5•S!

V
S .ABC
=
1
3
SA.S
ABC
=
1
3
.3a.a
2
= a
3
:75 ;*!!!
‰;!?O!AB{!

IC
IA

=
CD
AB
=
1
2
⇒
d (I ;(SCD))
d (A;(SCD ))
=
IC
AC
=
1
3
*!
p@!#c!5We3L!LBA!5<1!EcF!bE!343!

CD ⊥ (SAD )
*!+ˆ!Ez!5We3L!
LBA!5<1!bc!.d1!z!.$@!

AH ⊥ (SCD )
*!
!
!
X!
?O'!L1-A!5We3L!bEc!AB{!

1

AH
2
=
1
SA
2
+
1
AD
2
=
1
9a
2
+
1
a
2
⇒ AH =
3a
10
*!!!
p@!5•S!

d (I ;(SCD)) =
1
3
d (A;(SCD )) =
AH
3

=
a
10
*!!
Bh=)-)i_c*)>%d6jF)?D%3L!a$e3L!L1O3!5<1!.DkA!.%d!7\!lQSm!A$%!$O1!71>'!E:"R"R=;!5&!G:"Rn"R";F!
7Jo3L!.$f3L!

d :
x −1
1
=
y −1
1
=
z +1
1
*!p12.!I$JK3L!.D@3$!'P.!I$f3L!:q;!A$_O!EG!5&!(%3L!(%3L!
5<1!]*!?M3$!a$%,3L!A-A$!.j!]!723!'P.!I$f3L!:q;*!!
?O!AB{!

AB
! "!!
= (0;−2;−1)
F!7Jo3L!.$f3L!]!AB!5‹A!.K!A$€!I$JK3L!

u
!
= (1;1;1)
*!
uP.!I$f3L!:q;!A$_O!EG!5&!(%3L!(%3L!5<1!]!343 !AB !5.I .!h& !

n
!
= u
!
,AB
" !""
⎡
⎣
⎢
⎤
⎦
⎥
= (1;1;−2)
*!
bWS!DO!

(P ) : x + y − 2z + 2 = 0
*!
Œ^S!71>'!#:"R"Rn";!.$W\A!]!a$1!7B!

d (d ;(P )) = d (C ;(P )) =
1+1+ 2+ 2
1
2
+1
2
+ 2
2
= 6

*!!
nU%)#;@)op&)#4f6F!?@'!71>'!u!.$W\A!:q;!(O%!A$%!.O'!L1-A!EuG!7 • W*!
|•({!

M 1−
5
2 2
;
1
2 2
;
3
2
−
1
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
;M 1+
5

2 2
;−
1
2 2
;
3
2
+
1
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
*!!!!!
Bh=),)i_c*)>%d6jF!?D%3L!'P.!I$f3L!5<1!.DkA!.%d!7\!lQS!A$%!7Jo3L!.$f3L!

d : 6x + 8 y +11 = 0
*!p12.!
I$JK3L!.D@3$!7Jo3L!.Dr3!:?;!AB!.N'!g!0-3!aM3$!0s3L!

2

!5&!At.!lQF!lS Fu Z !hv3!hJw.!.$`%!A-A!7% d3!
.$f3L!EGF#cFuZ!.$%,!'T3!

S
IMN
= 1;AB = CD (x
I
> 0)
*!
?$`%!L1,!.$12.{!:?;!AB!0-3!aM3$!

R = 2
*!?O!AB{!

AB = 2 R
2
− d
2
(I;Ox);CD = 2 R
2
− d
2
(I;Oy)
*!!
c%!

AB = CD ⇒ d(I;Ox ) = d(I;Oy) ⇒
I (a;a)
I (a;−a)
⎡

⎣
⎢
⎢
(a > 0)
*!!
‰;!Z2W!

I (a;a) ⇒ d (I ;d) =
14a +11
10
;MN = 2 R
2
− d
2
(I ;d ) = 2 2− d
2
(I ;d )
*!
?O!AB{!
!

S
IMN
=
1
2
MN.d (I;d) = d(I ;d). 2− d
2
(I;d ) =1 ⇔ d (I ;d) =1
⇔

14a +11
10
= 1 ⇔ 14a +11 = 10(Vô nghiem do a > 0)
*!
‰;!Z2W!

I (a;−a) ⇒ d(I;d) =
11− 2a
10
;MN = 2 R
2
− d
2
(I ;d )
*!
?O!AB{!
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&)!
Ž!
!

S
IMN
=
1
2
MN .d (I ;d) = d (I ;d ). 2−d
2
(I ;d ) = 1 ⇔ d(I;d) =1

⇔
11−2a
10
= 1 ⇔ 11−2a = 10 ⇔
a =
1
2
a =
21
2
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⇒
I
1
2
;−
1
2
⎛
⎝
⎜
⎜
⎜

⎜
⎞
⎠
⎟
⎟
⎟
⎟
I
21
2
;−
21
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎡
⎣
⎢
⎢
⎢

⎢
⎢
⎢
⎢
*!
p•S!AB!=!7Jo3L!.Dr3!.$%,!'T3!0&1!.%-3!h&!

(T ) : x −
1
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
2
+ y +
1
2
⎛
⎝
⎜
⎜

⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
2
= 2;(T ) : x −
21
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
2
+ y +
21
2
⎛
⎝

⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
2
= 2
*!!
B4b)3F!#B!.$>!.$OS!L1,!.$12.!

S
IMN
= 1
!0s3L!L1,!.$12.!

S
IMN
!7d.!L1-!.D9!h<3!3$^.F!5&!"!h&!L1-!.D9!h<3!3$^.*!
?$•.!5•SF!.O!AB{!


S
IMN
= d(I ;d ). 2−d
2

(I ;d) ≤
d
2
(I ;d) + 2− d
2
(I ;d)
2
= 1
.
Dấu bằng xảy ra khi và chỉ khi

d (I ;d ) = 2−d
2
(I ;d ) ⇔ d(I ;d) =1
. !!
Bh=)^)i_c*)>%d6jF!H1,1!$x!I$JK3L!.D@3$!

x(x + (x − y)(x − 2y)) = y( y + 3)
2
12x + y +7( y(x − y) + y(x − 2y)) = 8
( y + 3)
2
y
⎧
⎨
⎪
⎪
⎪
⎪
⎩

⎪
⎪
⎪
⎪
*!
|1•W!a1x3{!

y > 0 ⇒ x ≥ 2y > 0
{!
Z$N3!.$4'!S!5&%!$O1!52!AiO!I$JK3L!.D@3$!.$_!$O1!AiO!$x!(OW!7B!(%!(-3$!$O1!I$JK3L!.D@3$!.O!
7JwA{!

8x
2
+ 8x (x − y)(x −2y) =12xy + y
2
+ 7y( y(x − y) + y(x −2y))
⇔ 8.
x
2
y
2
+ 8
x
y
x
y
−1
⎛
⎝

⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
x
y
−2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
=
12x
y
+1+ 7
x

y
−1 +
x
y
−2
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
*!
|P.!

t =
x
y
≥ 2
I$JK3L!.D@3$!.D•!.$&3${!!!

8t
2

+ 8t (t −1)(t − 2) = 12t +1+ 7( t −1 + t − 2)
*!
|P.!

a = t −1 + t − 2 = 2t −3+ 2 (t −1)(t −2) ≥ 2t − 3 ≥1
R!5&!
!

a
2
= 2t −3+ 2 (t −1)(t − 2) ⇒ 2 (t −1)(t − 2) = a
2
− 2t + 3
*!
q$JK3L!.D@3$!.D•!.$&3${!
!

8t
2
+ 4t (a
2
− 2t + 3) =12t +1+ 7a ⇔ 4ta
2
= 1+ 7a ⇔ 4t =
1+7a
a
2
*!
?O!AB!

4t ≥ 8 ⇒
1+7a
a
2
≥ 8 ⇔ 8a
2
−7a −1 ≤ 0 ⇔ −
1
8
≤ a ≤1
*!
uP.!a$-A!

a ≥1 ⇒ a = 1 ⇔ t = 2 ⇔ x = 2 y
*!?$OS!5&%!I$JK3L!.D@3$!7vW!AiO!$x!.O!7JwA{!!!!
!

4y
2
= y ( y + 3) ⇔ y 4y y − y −3
⎡
⎣
⎢
⎤
⎦
⎥
= 0
⇔ y ( y −1)(4 y + 3 y + 3) = 0 ⇔ y =1(y > 0) ⇒ x = 2
*!
p•S!$x!I$JK3L!.D@3$!AB!3L$1x'!]WS!3$^.!

(x; y) = (2;1)
*!!!
!
!
•!
B4b)3F!|>!L1,1!I$JK3L!.D@3$!

8t
2
+ 8t (t −1)(t − 2) = 12t +1+ 7( t −1 + t − 2)
.O!AB!.$>!(†!
]k3L!I$JK3L!I$-I!$&'!()!$%PA!h143!$wI*!!
Bh=)+)i_c*)>%d6jF!#$%!QFSFm!h&!A-A!()!.$/A!.$%,!'T3!

x
2
+ y
2
+ z
2
= 2
*!?@'!L1-!.D9!3$y!3$^.!AiO!
01>W!.$_A!

P =
1
x − y + 2
+
1

y − z + 2
+
1
z − x + 2
+
1
xy + yz + zx + 2
*!
c%!71•W!a1x3!5&!q!7)1!Q_3L!0O!0123!QFSFm!343!a$e3L!'^.!.M3$!.Š3L!~W !L1,!(†!

x ≥ y ≥ z
*!
+$1!7B!!
!

P =
1
x − y + 2
+
1
y − z + 2
+
1
x − z + 2
+
1
xy + yz + zx + 2
*!
b†!]k3L!0^.!7f3L!.$_A!Eu!‘Hu!]d3L!A\3L!'VW!()!.O!AB{!
!

1
x − y + 2
+
1
y − z + 2
≥
4
x − z + 4
*!
b†!]k3L!0^.!7f3L!.$_A!#OWA$S!‘bA$’ODm!.O!AB{!
!

(x − z)
2
= (x − y) + ( y − z)
⎡
⎣
⎢
⎤
⎦
⎥
2
≤ 2 (x − y)
2
+ (y − z)
2
⎡
⎣
⎢

⎤
⎦
⎥
*!
bWS!DO!!

3
2
(x − z)
2
≤ (x − y)
2
+ (y − z)
2
+ (z − x )
2
= 4− 2(xy + yz + zx)
⇒ xy + yz + zx + 2 ≤ 4−
3
4
(x − z)
2
*!
p@!5•S!

P ≥
4
x − z + 4
+
1

x − z + 2
+
2
16−3(x − z )
2
*!
|P.!

t = x − z ∈ 0;2
⎡
⎣
⎢
⎤
⎦
⎥
F!0•1!5@!

x − z ≤ 2(x
2
+ z
2
) ≤ 2
*!
p@!5•S!

P ≥ f (t ) =
4
t + 4
+
1

t + 2
+
2
16−3t
2
*!
U‹.!$&'!()!

f (t ) =
4
t + 4
+
1
t + 2
+
2
16−3t
2
.D43!7%d3!“”R=•!.O!AB!
!

f '(t) =
3t
(16−3t
2
)
3
−
1
(t + 2)

2
−
4
(t + 4)
2
≤
6
64
−
1
16
−
4
36
< 0
*!
bWS!DO!–:.;!3L$9A$!0123!.D43!7%d3!“”R=•*!p@!5•S!

P ≥ f (t ) ≥ f (2) =
23
12
*!
c^W!0s3L!7d.!.d1!

x = 1; y = 0; z = −1
*!
p•S!L1-!.D9!3$y!3$^.!AiO!q!0s3L!=C•"=!7d.!.d1!

(x; y; z) = (1;0;−1)
$%PA!A-A!$%-3!59*!

B4b)3F!p<1!0&1!.%-3!A$_O!]^W!L1-!.D9!.WSx.!7)1!.J!]WS!0O3!7vW!h&!I$-!]^W!.D9!.WSx.!7)1!:71•W!
3&S!QW^.!I$ !.j!.M3$!7)1!Q_3L!AiO!0&1!.%-3!$%PA!]/O!5&%!71•W!a1x3!L1,!.$12.!A$%!(‡3;*!p1xA!
7-3$!L1-!:Qnm;!.$`%!A-A!7d1!hJw3L!7)1!Q_3L!]/O!5&%!71•W!a1x3!0&1!.%-3!5&!7-3$!L1-!$OS!(†!
]k3L{!


(x − z)
2
= (x − y) + ( y − z)
⎡
⎣
⎢
⎤
⎦
⎥
2
≤ 2 (x − y)
2
+ (y − z)
2
⎡
⎣
⎢
⎤
⎦
⎥
.
!!!!!
!!!!!!!!
!!

Khoá giải đề THPT Quốc Gia Môn Toán – Thầy Đặng Thành Nam – Mathlinks.vn!
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—!
!!
!!!
!!!
!