!"#$%&'('%)*%+",+% /0%&'1%+"23%)45&%+"65"%517%
"89:;<=>%?@AB%CBB%C?C%% )D<E%FG%<HIJ%K%HLM%N;<H%<HO<%PQ%RS;%HLM%THU%%%
0H;%9;V9>%7W9H:;<FNXY<!
"!
!H8Z%E;[;%R\%+",+%-Q]M%&;W%^%+H_`>%)a<E%+Hb<H%5WJ%
7c<>%+8Z<d%)*%e/%CKfg?%
5Eb`%9H;%>%h?f?ifC?hg%
+Hj;%E;W<%:bJ%kb;>%hl?%THm9n%FHc<E%Fo%9Hj;%E;W<%E;W8%R\%
p;q<%Hr%RD<E%FG%FH8Z%HLM%^%"89:;<=>%?@AB%CBB%C?C%^%0H;%9;V9>%sssXJW9H:;<FNXY<%
0tQ%h%uCn?%R;oJvX%#$%!$&'!()!

y =
2x
x− 2
(1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* #$%!71='!>:<?<;*!@A'!71='!B!.$CDE!:";!(F%!E$%!.12G!.12G!.CH23!EIF!:";!.J1!B!5CK3L!LME!5N1!
7OP3L!.$Q3L!>B*!
0tQ%C%uhn?%R;oJvX%
F; R1,1!G$OS3L!.TA3$!

tanx −1= 2 sin(x +
3π
4
)
*!!
0; #$%!()!G$UE!V!.$%,!'W3!

(1+ 2i).z = 3+ i
*!@X3$!L1-!.T9!EIF!01=C!.$UE!

A = z
4
− z
2
+1
*!
0tQ%K%u?ng%R;oJvX!R1,1!G$OS3L!.TA3$!

log
2
x +1
x
= log
4
x
2
+1
*!
0tQ%i%uhn?%R;oJvX!R1,1!0Y.!G$OS3L!.TA3$!

x +1+ x
2
+ x +1 −
x
2
− x +1+1
x
≤
1

x
− x −1
*!
0tQ%g%uhn?%R;oJvX!@X3$!.XE$!G$Z3!

I =
(x +1)
2
− x.ln x
x
2
dx
2
3
∫
*!!
0tQ%B%uhn?%R;oJvX!#$%!$A3$![\3L!.T]!7U3L!^_#*^`_`#`!EM!7-H!^_#![&!.F'!L1-E!7aC!EJ3$!Fb!

A 'B =
a 7
2
*!Rc1!Bbdbe![f3![Og.![&!.TC3L!71='!E-E!EJ3$!^`_`b^`#`b##`*!@X3$!.$=!.XE$!h$)1![\3L!
.T]!^_#*^`_`#`!5&!h$%,3L!E-E$!.i!71='!^!723!'j.!G$Q3L!:Bde;*!!
0tQ%A%uhn?%R;oJvX!@T%3L!'j.!G$Q3L!5N1!.T]E!.%J!7D!klH!E$%!7OP3L!.Tm3!

(C ) : x
2
+ y
2
= 1

*!Rc1!
^![&!71='!.$CDE!7OP3L!.$Q3L!

y −3 = 0
!5&!_b#![f3![Og.![&!E-E!.12G!71='!EIF!.12G!.CH23!hn!.i!^!
723!:#;*!oCF!_!hn!7OP3L!.$Q3L!(%3L!(%3L!5N1!^#!Ep.!:#;!.J1!qb!^q!Ep.!:#;!.J1!r*!@A'!.%J!7D!
71='!^b!012.!_r!Ep.!^#!.J1!71='!>:"?<;*!
0tQ%l%uhn?%R;oJvX%@T%3L!h$K3L!L1F3!5N1!$s!.T]E!.%J!7D!klHV!E$%!$F1!71='!^:"?<?t";!5&!_:u?u?<;*!
@A'!.%J!7D!71='!>!.$%,!'W3!

IA
!"!
= 2IB
!"!
*!v12.!G$OS3L!.TA3$!7OP3L!.$Q3L!

Δ
71!wCF!>b!Ep.!.T]E!kV!
5&!5CK3L!LME!5N1!7OP3L!.$Q3L!^_*!!!
0tQ%@%u?ng%R;oJvX%@i!'D.!3LZ3!$&3L!<u!EZC!$x1b!.T%3L!7M!EM!y!EZC!$x1!h$M!3LOP1!.F!lZH!z/3L!
.$&3$!$F1!7a!.$1!'{1!7a!.$1!L8'!"u!EZCb!5&!E-E!EZC!.T%3L!'D.!7a!7OgE!7-3$!()!.$U!./!.i!#ZC!"!
723!#ZC!"u*!@X3$!l-E!(CY.!7=!lZH!z/3L!7OgE!$F1!7a!.$1!'&!'{1!7a!.$1!7aC!L8'!<!EZC!$x1!h$M*!!
0tQ%h?%uhn?%R;oJvX!#$%!Fb0bEb'b3bG![&!E-E!()!.$/E!.$%,!'W3!

a
2
+ b
2
+ c
2

= m
2
+ n
2
+ p
2
= 9
*!@A'!
L1-!.T9![N3!3$Y.!EIF!01=C!.$UE!

P = 9− a− 2b−2c + 9− m− 2n− 2p + 9− am− bn− cp
*!!
www"x+www%
!"#$%&'('%)*%+",+% /0%&'1%+"23%)45&%+"65"%517%
"89:;<=>%?@AB%CBB%C?C%% )D<E%FG%<HIJ%K%HLM%N;<H%<HO<%PQ%RS;%HLM%THU%%%
0H;%9;V9>%7W9H:;<FNXY<!
<!
%
,"y5%+z0"%{|5"%p.}5%)$,%$5%
0tQ%h%uCn?%R;oJvX%#$%!$&'!()!

y =
2x
x− 2
(1)
*!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* #$%!71='!>:<?<;*!@A'!71='!B!.$CDE!:";!(F%!E$%!.12G!.12G!.CH23!EIF!:";!.J1!B!5CK3L!LME!5N1!
7OP3L!.$Q3L!>B*!
"* |cE!(13$!./!L1,1*!

<* Rc1!

M(m;
2m
m−2
)∈ (1),m ≠ 2
b!(CH!TF!$s!()!LME!EIF!7OP3L!.$Q3L!>B![&!

k
IM
=
y
M
− y
I
x
M
− x
I
=
2m
m−2
− 2
m−2
=
4
(m− 2)
2
*!
};!|s!()!LME!EIF!.12G!.CH23!5N1!:";!.J1!B![&!


k = y'(m) =
−4
(m− 2)
2
*!
@$~%!L1,!.$12.!.F!EM•!

k
IM
.k = −1⇔
−16
(m− 2)
4
= −1⇔ (m −2)
4
= 16 ⇔
m−2 = 2
m−2 = −2
⎡
⎣
⎢
⎢
⎢
⇔
m = 4
m = 0
⎡
⎣
⎢

⎢
⎢
*!
};!v€H!EM!$F1!71='!B:y?y;!5&!B:u?u;*!!!!!
0tQ%C%uhn?%R;oJvX%
F; R1,1!G$OS3L!.TA3$!

tanx −1= 2 sin(x +
3π
4
)
*!!
0; #$%!()!G$UE!V!.$%,!'W3!

(1+ 2i).z = 3+ i
*!@X3$!L1-!.T9!EIF!01=C!.$UE!

A = z
4
− z
2
+1
*!
F; •1aC!h1s3•!

cosx ≠ 0 ⇔ x ≠
π
2
+ kπ,k ∈ !
*!

e$OS3L!.TA3$!.OS3L!7OS3L!5N1•!

sin x− cosx
cosx
= −sin x + cosx ⇔ (sin x−cosx)(cosx +1) = 0
⇔
sin x = cosx
cosx = −1
⎡
⎣
⎢
⎢
⎢
⇔
tanx = 1
cosx = −1
⎡
⎣
⎢
⎢
⎢
⇔
x =
π
4
+ kπ
x = π + k2π
⎡
⎣
⎢

⎢
⎢
⎢
,k ∈ !
*!
v€H!G$OS3L!.TA3$!EM!3L$1s'!

x =
π
4
+ kπ;x = π + k2π,k ∈ !
*!!
0; !@F!EM•!

z =
3+ i
1+ 2i
=
(3+ i)(1− 2i)
5
=
5−5i
5
= 1− i ⇒ z = 2
*!
q%!7M!

A = ( 2)
4
− ( 2)

2
+1= 4− 2+1= 3
*!!
0tQ%K%u?ng%R;oJvX!R1,1!G$OS3L!.TA3$!

log
2
(
x +1
x
)= log
4
x
2
+1
*!
!"#$%&'('%)*%+",+% /0%&'1%+"23%)45&%+"65"%517%
"89:;<=>%?@AB%CBB%C?C%% )D<E%FG%<HIJ%K%HLM%N;<H%<HO<%PQ%RS;%HLM%THU%%%
0H;%9;V9>%7W9H:;<FNXY<!
‚!
•1aC!h1s3•!

x ≠ 0
x +1
x
> 0
⎧
⎨
⎪
⎪

⎪
⎪
⎩
⎪
⎪
⎪
⎪
⇔
x > 0
x <−1
⎡
⎣
⎢
⎢
⎢
*!e$OS3L!.TA3$!.OS3L!7OS3L!5N1•!

log
2
(
x +1
x
)= log
2
x + log
2
2 ⇔ log
2
(
x +1

x
)= log
2
2 x ⇔
x +1
x
= 2 x ⇔
x > 0
2x =
x +1
x
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
x <−1
−2x =
x +1
x
⎧
⎨
⎪
⎪

⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⇔ x = 1
*!
v€H!G$OS3L!.TA3$!EM!3L$1s'!zCH!3$Y.!

x = 1
*!!
0tQ%i%uhn?%R;oJvX!R1,1!0Y.!G$OS3L!.TA3$!

x +1+ x
2

+ x +1 −
x
2
− x +1+1
x
≤
1
x
− x −1
*!
•1aC!h1s3•!

x > 0
*!
_Y.!G$OS3L!.TA3$!.OS3L!7OS3L!5N1•!!!

x +1+ x
2
+ x +1 −
x
2
− x +1+1
x
≤
1
x
− x −1
(x +1)+ (x +1)
2
−(x +1)+1 + x +1≤

1
x
+
1
x
2
−
1
x
+1 +
1
x
(1)
*!
ƒ„.!$&'!()!

f (t)= t+ t
2
− t +1 + t
b!.F!EM•!

f '(t) = 1+
1+
2t −1
2 t
2
− t +1
2 t + t
2
− t +1

= 1+
2t −1+ 2 t
2
− t +1
4 t + t
2
− t +1. t
2
− t +1
> 0
b!!
0…1!5A!

2t −1+ 2 t
2
−t +1 = 4t
2
− 4t + 4 + 2t −1= (2t −1)
2
+ 3 + 2t −1> 2t −1 + 2t −1≥ 0
*!!!!
q%!7M!†:.;![&!$&'!783L!0123*!vA!5€H

(1) ⇔ x +1≤
1
x
⇔
x
2
+ x −1

x
≤ 0 ⇔ 0< x ≤
−1+ 5
2
(do x > 0)
*!!
v€H!.€G!3L$1s'!EIF!0Y.!G$OS3L!.TA3$![&!

S = 0;
−1+ 5
2
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎤
⎦
⎥
⎥
⎥
*!!
{b;%9OT%9P~<E%9•w%%
R1,1!0Y.!G$OS3L!.TA3$!

x +1+ x
2
+ x +1 − 2x + 4x

2
− 2x +1 ≥ x −1
*!•‡(•!

S = 1;+∞
⎡
⎣
⎢
)
*!!!
0tQ%g%uhn?%R;oJvX!@X3$!.XE$!G$Z3!

I =
(x +1)
2
− x.ln x
x
2
dx
2
3
∫
*!!
};!@F!EM•!!

I =
(x +1)
2
x
2

dx
2
3
∫
−
ln x
x
dx
2
3
∫
*!
!"#$%&'('%)*%+",+% /0%&'1%+"23%)45&%+"65"%517%
"89:;<=>%?@AB%CBB%C?C%% )D<E%FG%<HIJ%K%HLM%N;<H%<HO<%PQ%RS;%HLM%THU%%%
0H;%9;V9>%7W9H:;<FNXY<!
y!
};!

I
1
=
(x +1)
2
x
2
dx
2
3
∫
= (1+

2
x
+
1
x
2
)dx
2
3
∫
= (x + 2ln x −
1
x
)
3
2
=
7
6
+ 2ln
3
2
*!
};!

I
2
=
ln x
x

dx
2
3
∫
= ln xd(lnx)
2
3
∫
=
1
2
ln
2
x
3
2
=
ln
2
3−ln
2
2
2
*!
};!

I = I
1
− I
2

=
7
6
+ 2ln
3
2
−
ln
2
3− ln
2
2
2
*!!!!
0tQ%B%uhn?%R;oJvX!#$%!$A3$![\3L!.T]!7U3L!^_#*^`_`#`!EM!7-H!^_#![&!.F'!L1-E!7aC!EJ3$!Fb!

A 'B =
a 7
2
*!Rc1!Bbdbe![f3![Og.![&!.TC3L!71='!E-E!EJ3$!^`_`b^`#`b##`*!@X3$!.$=!.XE$!h$)1![\3L!
.T]!^_#*^`_`#`!5&!h$%,3L!E-E$!.i!71='!^!723!'j.!G$Q3L!:Bde;*!!
!
};!

S
ABC
=
1
2
AB.AC sin60

0
=
a
2
3
4
*!!!
@F'!L1-E!5CK3L!^`^_!EM•!
!

AA'= A 'B
2
− AB
2
=
7a
2
4
− a
2
=
a 3
2
*!!
};!

V
ABC.A 'B 'C '
= AA '.S
ABC

=
a 3
2
.
a
2
3
4
=
3a
3
8
:75 ;*!
};!Rc1!e`![&!.TC3L!71='!EIF!__`!.F!EM!ee`‡‡_`#`‡‡Bd!343!
'j.!G$Q3L!:Bde;![&!'j.!G$Q3L!:Bdee`;*!
Rc1!rbr`bˆb‰![f3![Og.![&!.TC3L!71='!EIF!_`#`b_#bBdbrr`!.F!
EM•!ˆ‰![&!L1F%!.CH23!EIF!'j.!G$Q3L!:^^`r`r;!5N1!'j.!
G$Q3L!:Bde;*!
•OP3L!.$Q3L!^r`!Ep.!ˆ‰!.J1!|b!Ep.!^`r!.J1!>*!
@F!EM•!_#!5CK3L!LME!5N1!'j.!G$Q3L!:^^`r`r;!5&!ee`‡‡_#!343!ee`!5CK3L!LME!5N1!^>!:";*!
Bj.!h$-E!.F'!L1-E!^^`r!5CK3L!EZ3!.J1!^!5A!

AE = AA '=
a 3
2
b!z%!7M!^>!5CK3L!LME!5N1!^`r!:<;*!!
Bj.!h$-E!^`r‡‡ˆ‰!:‚;*!
@i!:";b:<;b:‚;!(CH!TF•!

AI ⊥ (MNP) ⇒ AH = d(A;(MNP))

*!
};!q%!ˆ‰![&!7OP3L!.TC3L!0A3$!EIF!.F'!L1-E!^`r`r!343!|![&!.TC3L!71='!EIF!>r`b![J1!EM!>![&!.TC3L!
71='!EIF!^r`!343!

AH =
3
4
AE ' =
3
4
AE
2
+ EE '
2
=
3
4
3a
2
4
+
3a
2
4
=
3a 6
8
*!
v€H!


d(A;(MNP))=
3a 6
8
*!
0ZMH%C>!!Rc1!r![&!.TC3L!71='!_#b!E$c3!$s!.T]E!.%J!7D!klHV!(F%!E$%•!

A(0;0;0),E(
a 3
2
;0;0),B(
a 3
2
;−
a
2
;0),C(
a 3
2
;
a
2
;0),A'(0;0;
a 3
2
),B '(
a 3
2
;−
a
2

;
a 3
2
),C '(
a 3
2
;
a
2
;
a 3
2
)
*!
+$1!7M!.%J!7D!Bbdbe![f3![Og.![&!

M(
a 3
4
;−
a
4
;
a 3
2
),N(
a 3
4
;
a

4
;
a 3
2
),P(
a 3
2
;
a
2
;
a 3
4
)
*!
KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%THẦY%ĐẶNG%THÀNH%NAM%
Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%%
Chi%tiết:%Mathlinks.vn!
"!
#$%!&'(!!!!

MN
! "!!
= (0;
a
2
;0)/ /(0;1;0),MP
! "!!
= (
a 3

4
;
3a
4
;−
a 3
4
) / /(1; 3;−1)
)!*+!*,-!./!0120!.$%34!-5'!
6789:!;+!

n
(MNP)
! "!!!!
=
1 0
3 −1
;
0 0
−1 1
;
0 1
1 3
⎛
⎝
⎜
⎜
⎜
⎜
⎜

⎞
⎠
⎟
⎟
⎟
⎟
⎟
⎟
= (−1;0;−1)
<!!
=>!*?%!01@/4A!.&>41!BC.!01D4A!6789:!;+!

x + z−
3a 3
4
= 0
<!
E'!-F(!

d(A;(MNP))=
−
3a 3
4
2
=
3a 6
8
<!!!!
Câu%7%(1,0%điểm).!E&G4A!BC.!01D4A!*HI!.&J-!.GK!LM!NO%!-1G!L@P4A!.&Q4!


(C ) : x
2
+ y
2
= 1
<!RSI!
T!;+!LIUB!.1$M-!L@P4A!.1D4A!

y −3 = 0
!*+!V)W!;X4!;@Y.!;+!-2-!.I30!LIUB!-5'!.I30!.$%34!Z[!.\!T!
L34!6W:<!]$'!V!Z[!L@P4A!.1D4A!^G4A!^G4A!*HI!TW!-_.!6W:!.KI!`)!T`!-_.!6W:!.KI!a<!E>B!.GK!LM!
LIUB!T)!bI3.!Va!-_.!TW!.KI!LIUB!c6def:<!
!
g@P4A!.&Q4!6W:!-F!.hB!;+!Ai-!.GK!LM!N6jej:)!b24!Zk41!
bl4A!d<!
Phát%hiện%tính%chất%hình%học:%
E'!-F!c!;+!.&$4A!LIUB!TW!
Chứng&minh.&
cW!;+!.I30!.$%34!-5'!L@P4A!.&Q4!6W:!4m4!

IC
2
= IE.IB (1)
<!
E1nG!.k41!-1o.!AF-!4pq!.I30!*+!AF-!.KG!brI!.I30!.$%34!*+!
sh%!-$4A!.'!-F(!
!

EBA
!

= BDA
!
BDA
!
= EAI
!
(so le)
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇒ EBA
!
= EAI
!
<!
`G!LF!.'B!AI2-!cVT!Lt4A!sK4A!*HI!.'B!AI2-!cTa6A<A:<!!
u:!#$%!&'(!

IA
IE
=
IB
IA
⇒ IA

2
= IB.IE (2)
<!
E\!6d:!*+!6f:!^$%!&'(!

IA = IC
!)!1'%!c!;+!.&$4A!LIUB!TW<!
u:!RSI!T6'ev:!.1$M-!L@P4A!.1D4A!

y −3 = 0
)!sG!c!;+!.&$4A!LIUB!TW!4m4!W6fw'ed:<!
7C.!Z12-(!

OC
2
= 1⇔ (2− a)
2
+1=1 ⇔ a = 2 ⇒ A(2;3)
<!
Kết%luận:!=?%!T6fev:!;+!LIUB!-X4!.>B<!!!
Bình%luận:!E'!-F!.1U!-1x4A!BI41!c!;+!.&$4A!LIUB!TW!41@!^'$(!
y[!.I30!.$%34!!*HI!6W:!.KI!`)!-_.!TW!.KI!7<!z>41!.1'4A!TV`7!-F!

DBA
!
= BDM
!
!4m4!;+!1>41!
.1'4A!-h4<!#$%!&'(!


BAI
!
= M
"
*+!W!;+!.&$4A!LIUB!-5'!T7<!
RSI!y!;+!.&$4A!LIUB!-5'!T`)!.'!-F!Wy!;+!L@P4A!.&$4A!b>41!-5'!.'B!AI2-!T`7!4m4!
!

CK =
MD
2
=
AB
2
=
AC
2
<!
E'B!AI2-!TVc!*+!WTy!-F!

AB = AC ,ABI
!
= CAK
!
, BAI
!
= ACK
!
4m4!bl4A!41'$<!
KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%THẦY%ĐẶNG%THÀNH%NAM%

Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%%
Chi%tiết:%Mathlinks.vn!
{!
`G!LF!

AI = CK
)!*>!*?%!

AI =
AC
2
⇒ I
!;+!.&$4A!LIUB!-5'!TW<!
Câu%8%(1,0%điểm).%E&G4A!Z1p4A!AI'4!*HI!1|!.&J-!.GK!LM!NO%}!-1G!1'I!LIUB!T6defewd:!*+!V6jejef:<!
E>B!.GK!LM!LIUB!c!.1G~!B•4!

IA
!"!
= 2IB
!"!
<!=I3.!01@/4A!.&>41!L@P4A!.1D4A!

Δ
LI!€$'!c)!-_.!.&J-!N}!
*+!*$p4A!AF-!*HI!L@P4A!.1D4A!TV<!!!
u:!=>!

IA
!"!
= 2IB

!"!
4m4!V!;+!.&$4A!LIUB!-5'!cT)!sG!LF!c6wdewfe":<!
u:!RI~!^•!L@P4A!.1D4A!

Δ
-_.!.&J-!N}!.KI!LIUB!76jeje.:)!.'!-F(!

IM
! "!
= (1;2;t − 5)
<!
g@P4A!.1D4A!TV!-F!*,-!./!-1‚!01@/4A!

AB
! "!
= (−1;−2;3)
<!
`G!TV!*+!!

Δ
*$p4A!AF-!4m4(!!

−1.1− 2.2+ 3(t − 5) = 0 ⇔ 3t −10= 0 ⇔ t =
10
3
⇒ IM
! "!
= (1;2;−
5
3

) / /(3;6;−5)
<!
u:!g@P4A!.1D4A!

Δ
LI!€$'!c!*+!-F!*,-!./!-1‚!01@/4A!6ve{ew":!4m4!-F!01@/4A!.&>41!;+(!

Δ :
x +1
3
=
y + 2
6
=
z− 5
−5
<!
Câu%9%(0,5%điểm).%E\!BM.!4Ah4!1+4A!fj!-h$!1ƒI)!.&G4A!LF!-F!„!-h$!1ƒI!Z1F!4A@PI!.'!Oh%!s…4A!
.1+41!1'I!L†!.1I!B‡I!L†!.1I!AtB!dj!-h$)!*+!-2-!-h$!.&G4A!BM.!L†!L@Y-!L241!^i!.1x!.…!.\!Wh$!d!
L34!Wh$!dj<!Ek41!O2-!^$o.!LU!Oh%!s…4A!L@Y-!1'I!L†!.1I!B+!B‡I!L†!.1I!L†$!AtB!f!-h$!1ƒI!Z1F<!
y1p4A!AI'4!Bˆ$!;+!^i!-2-1!Oh%!s…4A!1'I!L†!.1I!B‡I!L†!.1I!AtB!dj!-h$!L@Y-!-1S4!&'!.\!4Ah4!
1+4A!fj!-h$!1ƒI<!
u!W1S4!&'!dj!-h$!1ƒI!-1G!L†!.1x!41o.)!^'$!LF!^_0!O30!.1nG!.1x!.…!.\!-h$!d!L34!-h$!dj!-F!

C
20
10
.10!
!-2-1<!
u!dj!-h$!-Q4!;KI!;o%!;+B!L†!.1x!1'I)!*+!^_0!O30!.1nG!.1x!.…!.\!-h$!d!L34!-h$!dj!-F!dj‰!-2-1<!

=?%!Z1p4A!AI'4!Bˆ$!

Ω = C
20
10
.10!.10!= (10!)
2
.C
20
10
<!
u:!RSI!T!;+!bI34!-i!Oh%!s…4A!L@Y-!1'I!L†!.1I!B‡I!L†!AtB!f!-h$!1ƒI!Z1F<!
u!W1S4!&'!f!-h$!1ƒI!Z1F!.&G4A!„!-h$)!*+!Š!-h$!1ƒI!s‹!.&G4A!d{!-h$!-1G!L†!.1x!41o.)!^'$!LF!^_0!
O30!dj!-h$!4+%!.1nG!.1x!.…!.\!-h$!d!L34!-h$!dj!-F!

C
4
2
.C
16
8
.10!
-2-1<!
u!dj!-h$!-Q4!;KI!;o%!;+B!L†!.1x!1'I)!*+!^_0!O30!.1nG!.1x!.…!.\!-h$!d!L34!-h$!dj!-F!dj‰!-2-1<!
=?%!^i!01X4!.•!-5'!bI34!-i!T!;+!

Ω
A
= C
4

2
.C
16
8
.10!.10!= (10!)
2
.C
4
2
.C
16
8
<!
=?%!O2-!^$o.!-X4!.k41!

P(A)=
Ω
A
Ω
=
(10!)
2
.C
4
2
.C
16
8
(10!)
2

.C
20
10
=
135
323
<!!!!
Câu%10%(1,0%điểm).!W1G!')b)-)B)4)0!;+!-2-!^i!.1…-!.1G~!B•4!

a
2
+ b
2
+ c
2
= m
2
+ n
2
+ p
2
= 9
<!E>B!
AI2!.&q!;H4!41o.!-5'!bIU$!.1x-!

P = 9− a− 2b−2c + 9− m− 2n− 2p + 9− am− bn− cp
<!!
E'!-F(!

P =

(a−1)
2
+ (b− 2)
2
+ (c− 2)
2
2
+
(m−1)
2
+ (n− 2)
2
+ (p− 2)
2
2
+
(a− m)
2
+ (b− n)
2
+ (c− p)
2
2
<!
KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%THẦY%ĐẶNG%THÀNH%NAM%
Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%%
Chi%tiết:%Mathlinks.vn!
Œ!
u:!E&G4A!Z1p4A!AI'4!NO%})!O,.!b'!LIUB!T6'ebe-:)!V6Be4e0:)!W6defef:)!.'!-F!T)V)W!-•4A!.1$M-!
BC.!-X$!6#:!-F!01@/4A!.&>41(


x
2
+ y
2
+ z
2
= 9
)!b24!Zk41!

R = 3
<!!
=+!

P =
1
2
(AB + BC +CA)
)!.'!ASI!69:!;+!BC.!01D4A!-1x'!.'B!AI2-!TVW<!y1I!LF!69:!-_.!6#:!.1nG!
AI'G!.$%34!B+!BM.!L@P4A!.&Q4!b24!Zk41!

R '= R
2
− d
2
(O;(P)) ≤ R
<!!!
u:!E'!-F(!

AB + BC +CA ≤ 3 3R' ≤ 3 3R = 9 3 ⇒ P ≤

9 6
2
<!
`o$!bl4A!O~%!&'!Z1I!*+!-1‚!Z1I!.'B!AI2-!TVW!L†$!*+!N!;+!.&S4A!.hB!-5'!.'B!AI2-!TVW<!!
`G!;$p4!-F!BC.!01D4A!69:!LI!€$'!N)W!*+!69:!-_.!6#:!.1nG!AI'G!.$%34!;+!BM.!L@P4A!.&Q4!6W:)!
.&m4!6W:!;$p4!.>B!L@Y-!1'I!LIUB!T)V!LU!.'B!AI2-!TVW!4m4!so$!bl4A!O~%!&'<!!
=?%!AI2!.&q!;H4!41o.!-5'!9!bl4A!

9 6
2
<!!!!
Bình%luận:%WF!.1U!-J!.1U!1G2!so$!bl4A!41@!^'$(!

a
2
+ b
2
+ c
2
= 9
(a−1)
2
+ (b− 2)
2
+ (c− 2)
2
= 27
m
2
+ n

2
+ p
2
= 9
(m−1)
2
+ (n− 2)
2
+ (p− 2)
2
= 27
a + m+1
3
=
b + n+ 2
3
=
c+ p+ 2
3
= 0
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
<!
u:!ŽPI!AI~I!.&m4!L•!^•!b+I!.G24!-…-!.&q!1>41!1S-!^'$(!
Tam giác ABC nội tiếp đường tròn (O) cho trước có chu vi lớn nhất khi ABC là tam giác đều.
Chứng minh.
Gọi R là bán kính đường tròn, theo định lý hàm số Sin ta có:


BC = 2Rsin A,CA = 2RsinB,AB = 2RsinC
.
Vì vậy chu vi tam giác ABC là

P = 2R(sin A + sinB + sinC) (1)
,
Mặt khác:

sin A + sinB + sinC = 2sin
A + B

2
cos
A − B
2
+ 2sin
C
2
cos
C
2
≤ 2cos
C
2
1+ sin
C
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
(2)

.
#•!sJ4A!bo.!LD4A!.1x-!T7!•R7!-1G!b'!^i!.1…-!Z1p4A!hB!.'!-F(!

2cos
C
2
1+ sin
C
2
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
= 2. 2cos
2
C
2
(1+ sin
C
2
)(1+ sin

C
2
) ≤ 2.
2cos
2
C
2
+1+ sin
C
2
+1+ sin
C
2
3
⎛
⎝
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
3
= 2.
9
2
−2(sin
C
2
−
1
2
)
2
3
⎛
⎝
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
3
≤ 2. (
3
2
)
3
=
3 3
2
(3)
<!!

Từ (1),(2),(3) ta có:

P ≤ 3 3R
. Dấu bằng xảy ra khi và chỉ khi

A = B
sin
C
2
=
1
2
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⇔ A = B = C = 60
0
.
KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%THẦY%ĐẶNG%THÀNH%NAM%
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Š!

Cách&2:!#•!sJ4A!bo.!LD4A!.1x-!W'$-1%!•#-1•'&}!.'!-F!!

P =
1
2
(AB + BC +CA)≤
3
2
(AB
2
+ BC
2
+ CA
2
)
=
3
2
(OB
! "!
−OA
! "!
)
2
+ (OC
! "!!
−OB
! "!
)
2

+ (OA
! "!
−OC
! "!!
)
2
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
=
3
2
(6R
2
−2(OA
! "!
.OB
! "!
+ OC
! "!!
.OB
! "!
+ OA
! "!
.OC

! "!!
)
<!
u:!7C.!Z12-!ASI!R!;+!.&S4A!.hB!.'B!AI2-!TVW!.'!-F!
!

OA
! "!
+ OB
! "!
+ OC
! "!!
= 3OG
! "!!
⇒ (OA
! "!
+ OB
! "!
+ OC
! "!!
)
2
= 9OG
2
⇔ 3R
2
+ 2(OA
! "!
.OB
! "!

+ OC
! "!!
.OB
! "!
+ OA
! "!
.OC
! "!!
)= 9OG
2
⇒ 2(OA
! "!
.OB
! "!
+ OC
! "!!
.OB
! "!
+ OA
! "!
.OC
! "!!
)= 9OG
2
− 3R
2
<!
=>!*?%!

P ≤

3
2
(6R
2
−(9OG
2
−3R
2
)) =
27R
2
2
−
27OG
2
2
≤
27R
2
2
=
9 6
2
<!
`o$!bl4A!O~%!&'!Z1I!*+!-1‚!Z1I!

O ≡ G
AB = BC = CA
⎧
⎨

⎪
⎪
⎩
⎪
⎪
<!!!!
Bài%tập%tương%tự%w%
Bài số 01. Cho a,b,c,d là các số thực thoả mãn

a
2
+ b
2
= c
2
+ d
2
= 10
. Tìm giá trị lớn nhất của biểu thức

P = 10− a − 3b + 10−c −3d + 10− ac−bd
. Đ/s:

P
max
= 3 15
.
Bài số 02. Cho a,b,c,d là các số thực thoả mãn

a

2
+ b
2
= c
2
+ d
2
= 25
. Tìm giá trị lớn nhất của biểu thức

P = 25− 3a − 4b + 25−3c− 4d + 25− ac−bd
. Đ/s:

P
max
=
15 6
2
.
!