đề thi thử thpt quốc gia môn toán,đề số 18

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Khoá%giải%đề%THPT%Quốc%Gia%–%Thầy:%Đặng%Thành%Nam%
Môn:%Toán;%ĐỀ%SỐ%18/50%
Ngày%thi%:%22/03/2015%
Thời%gian%làm%bài:%180%phút,%không%kể%thời%gian%giao%đề%
Liên%hệ%đăng%ký%khoá%học%–%Hotline:%0976%266%202%–%Chi%tiết:%www.mathlinks.vn%%
Câu%1%(2,0%điểm).!#$%!$&'!()!

y = x
3
− x
2
+ 4 (1)
*!!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* =>'!.%?!7@!71A'!B!.$C@D!:";!(E%!D$%!.12F!.CG23!DHE!:";!.?1!B!.?%!5I1!$E1!.JKD!.%?!7@!'@.!
.E'!L1-D!DM3*!
Câu%2%(1,0%điểm).%
E; N1,1!F$OP3L!.J>3$!

2cos
2
x + 3 sin 2x = 2sin 5x +1
*!
0; N1,1!0Q.!F$OP3L!.J>3$!

log x.log(100x
2

) = 4
*!!!!
Câu%3%(1,0%điểm).!=R3$!.RD$!F$M3!

I =
1− x
2
ln x
x
dx
1
2
∫
*!
Câu%4%(1,0%điểm).%
E; S-D!793$!F$T3!,%!DHE!()!F$UD!

z = (3+ 2i )(−3+ i)+
1−3i
2− i
*!!
0; #$%!V$E1!.J1A3!

(1+ 2x )
n
= a
0
+ a
1
x + a

2
x
2
+ + a
n
x
n
*!=>'!$W!()!DHE!()!$?3L!D$UE!

x
8
!.J%3L!
V$E1!.J1A3!012.!

a
0
+ 9a
1
= 2a
2
+1
*!
Câu%5%(1,0%điểm).!#$%!$>3$!D$XF!Y*Z[#\!DX!7-G!Z[#\!]&!$>3$!.$%1^

AB = a 3,BD = 3a
*!_>3$!
D$12C!5C`3L!LXD!DHE!Y!.J43!'a.!F$b3L!:Z[#\;!.Jc3L!5I1!.JC3L!71A'!D?3$!Z#*![12.!D`(13!LXD!
.?%!0d1!'a.!F$b3L!:YZ[;!5&!'a.!F$b3L!:Z[#\;!0e3L!

21

7
*!=R3$!.$f%!E!.$A!.RD$!V$)1!D$XF!
Y*Z[#\!5&!0-3!VR3$!'a.!DTC!3L%?1!.12F!.U!g1W3!Y*Z#\*!!
Câu%6%(1,0%điểm).!=J%3L!V$`3L!L1E3!5I1!$W!.JKD!.%?!7@!hiGj!D$%!71A'!Z:"klkm";!5&!'a.!F$b3L!

(P ) : 2x − 2y + z + 6 = 0
*!n12.!F$OP3L!.J>3$!7Oo3L!.$b3L!g!71!pCE!Z!5&!5C`3L!LXD!5I1!'a.!
F$b3L!:q;*!=>'!.%?!7@!71A'!B!.$%,!'r3!

AM = 2
5&!V$%,3L!D-D$!.s!B!723!:q;!7?.!L1-!.J9!]I3!
3$Q.*!
Câu%7%(1,0%điểm).!=J%3L!'a.!F$b3L!5I1!.JKD!.%?!7@!hiG!D$%!.E'!L1-D!L1-D!Z[#!3@1!.12F!7Oo3L!
.Jt3!

(C ) : x
2
+ (y−
29
8
)
2
=
697
64
*!=>'!.%?!7@!7u3$!Z^!012.!7Oo3L!.Jt3!3L%?1!.12F!.E'!L1-D!v[#!Dw.!
Z[^Z#!]T3!]Ox.!.?1!D-D!71A'!

M (
21

10
;1), N (0;
13
5
),(M ≠ B, N ≠ C )
!5&!Z!DX!$%&3$!7@!M'*!
Câu%8%(1,0%điểm).!N1,1!$W!F$OP3L!.J>3$!

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

⎪
⎪
⎪
(x, y ∈ !)
*!
Câu%9%(1,0%điểm).%#$%!i^G^j!]&!D-D!()!F$UD!.$%,!'r3!

x = y = z = 1
*!=>'!L1-!.J9!]I3!3$Q.!DHE!
01AC!.$UD!

P = x + y − z + y + z − x + z + x − y
*!
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PHÂN%TÍCH%BÌNH%LUẬN%ĐÁP%ÁN%
Câu%1%(2,0%điểm).!#$%!$&'!()!

y = x
3
− x
2
+ 4 (1)
*!!
"* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";*!
<* =>'!.%?!7@!71A'!B!.$C@D!:";!(E%!D$%!.12F!.CG23!DHE!:";!.?1!B!.?%!5I1!$E1!.JKD!.%?!7@!'@.!
.E'!L1-D!DM3*!

"* _yD!(13$!./!L1,1*!
<* Ny1!

M (m;m
3
− m
2
+ 4)
]&!71A'!DT3!.>'*!
z;!=12F!.CG23!.?%!5I1!$E1!.JKD!.%?!7@!'@.!.E'!L1-D!DM3!343!.12F!.CG23!5C`3L!LXD!5I1!7Oo3L!
F$M3!L1-D!DHE!$E1!.JKD!.%?!7@k!!
z;!q$OP3L!.J>3$!$E1!7Oo3L!F$M3!L1-D!DHE!$E1!.JKD!.%?!7@!]&!

y = −x
$%aD!

y = x
*!
\%!7X!$W!()!LXD!DHE!.12F!.CG23!.?1!B!]&!

k = y'(m) = 3m
2
− 2m = ±1 ⇔
3m
2
− 2m −1= 0
3m
2
− 2m +1= 0
⎡

⎣
⎢
⎢
⎢
⇔
m =1
m = −
1
3
⎡
⎣
⎢
⎢
⎢
⎢
*!
Kết%luận:!#X!$E1!71A'!.$%,!'r3!]&!

M (1;3)
5&!

M (−
1
3
;
104
27
)
*!!!!!!!
Câu%2%(1,0%điểm).%

E; N1,1!F$OP3L!.J>3$!

2cos
2
x + 3 sin 2x = 2sin 5x +1
*!
0; N1,1!0Q.!F$OP3L!.J>3$!

log x.log(100x
2
) = 4
*!!!!
E; q$OP3L!.J>3$!.OP3L!7OP3L!5I1{!
!

cos2x +1+ 3 sin2x = 2sin5x +1
⇔ 3 sin 2x + cos2x = 2sin5x ⇔
3
2
sin2x +
1
2
cos2x = sin5x
⇔ sin(2x +
π
6
) = sin5x ⇔
5x = 2x +
π
6

+ k2π
5x =
5π
6
− 2x + k2π
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⇔
x =
π
18
+ k
2π
3
x =
5π
42
+ k
2π
7
⎡
⎣
⎢
⎢

⎢
⎢
⎢
⎢
*!
Kết%luận:!q$OP3L!.J>3$!DX!3L$1W'!]&!

x =
π
18
+ k
2π
3
;x =
5π
42
+ k
2π
7
,k ∈ !
*!!!
0; |1}C!V1W3{!

x > 0
*!
q$OP3L!.J>3$!.OP3L!7OP3L!5I1{!
!

log x.(log100+ log x
2

) = 4 ⇔ log x.(2+ 2log x ) = 4
⇔ log
2
x + log x −2 = 0 ⇔
log x = 1
log x = −2
⎡
⎣
⎢
⎢
⇔
x = 10
x =
1
100
⎡
⎣
⎢
⎢
⎢
⎢
*!
Kết%luận:!q$OP3L!.J>3$!DX!3L$1W'!

x = 10;x =
1
100
*!!!!
Câu%3%(1,0%điểm).!=R3$!.RD$!F$M3!

I =
1− x
2
ln x
x
dx
1
2
∫
*!
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~!
=E!DX{!!

I = (
1
x
− x ln x )dx
1
2
∫
= ln x
2
1
− x ln x dx
1
2
∫

= ln 2−K
*!
z;!|a.!

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

⎪
⎪
⎪
⎪
⇒ K =
x
2
2
ln x
2
1
−
xdx
2
1
2
∫
= 2ln 2−
x
2
4
2
1
= 2ln 2−
3
4
*!
Kết%luận:!n•G!

I = ln2− (2ln 2−

3
4
) =
3
4
− ln2
*!!!
Câu%4%(1,0%điểm).%
E; S-D!793$!F$T3!,%!DHE!()!F$UD!

z = (3+ 2i )(−3+ i)+
1−3i
2− i
*!!
0; #$%!V$E1!.J1A3!

(1+ 2x )
n
= a
0
+ a
1
x + a
2
x
2
+ + a
n
x
n

*!=>'!$W!()!DHE!()!$?3L!D$UE!

x
8
!.J%3L!
V$E1!.J1A3!012.!

a
0
+ 9a
1
= 2a
2
+1
*!
E; =E!DX{!

z = (−11−3i) +
(1− 3i )(2+ i )
5
= −11−3i +
5− 5i
5
= −11−3i +1− i = −10− 4i
*!
n•G!F$T3!,%!DHE!j!0e3L!m€*!
0; =E!DX{!

(1+ 2x )
n

= C
n
k
.2
k
.x
k
k=0
n
∑
= a
0
+ a
1
x + a
2
x
2
+ + a
n
x
n
⇒
a
0
= C
n
0
.2
0

= 1
a
1
= C
n
1
.2
1
= 2n
a
2
= C
n
2
.2
2
= 2n(n −1)
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪

*!
=$f%!L1,!.$12.!.E!DX{!
!

1+ 9.2n = 2n(n −1) +1 ⇔ 2n(n −10) = 0 ⇔ n = 10
*!
YCG!JE!$W!()!DHE!()!$?3L!D$UE!

x
8
!]&!

a
8
= C
10
8
.2
8
*!!!
Câu%5%(1,0%điểm).!#$%!$>3$!D$XF!Y*Z[#\!DX!7-G!Z[#\!]&!$>3$!.$%1^

AB = a 3,BD = 3a
*!_>3$!
D$12C!5C`3L!LXD!DHE!Y!.J43!'a.!F$b3L!:Z[#\;!.Jc3L!5I1!.JC3L!71A'!D?3$!Z#*![12.!D`(13!LXD!
.?%!0d1!'a.!F$b3L!:YZ[;!5&!'a.!F$b3L!:Z[#\;!0e3L!

21
7
*!=R3$!.$f%!E!.$A!.RD$!V$)1!D$XF!

Y*Z[#\!5&!0-3!VR3$!'a.!DTC!3L%?1!.12F!.U!g1W3!Y*Z#\*!!
Ny1!_!]&!L1E%!71A'!DHE!Z#!5&![\^!.E!DX!

SH ⊥ (ABCD )
*!
•F!gK3L!793$!]‚!$&'!()!D`(13!D$%!.E'!L1-D!Z[\!.E!DX{!
!

cos BAD
!
=
AB
2
+ AD
2
− BD
2
2AB.AD
=
3a
2
+ 3a
2
− 9a
2
2.a 3.a 3
= −
1
2
⇒ BAD

!
= 120
0
*!
\%!7X!D-D!.E'!L1-D!Z[#^!Z\#!7}C!D?3$!

a 3
*!
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Chi%tiết:%Mathlinks.vn!
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!
+ƒ!_„!5C`3L!LXD!5I1!Z[!.?1!„^!.E!DX{!

AB ⊥ HN
AB ⊥ SH
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⇒ AB ⊥ (SHN ) ⇒ ((SAB );(ABCD ))
!
= SNH
!
*!
=E'!L1-D!5C`3L!Z[_!DX{!

!

HN = AH .sin60
0
=
a 3
2
.
3
2
=
3a
4
*!
=E'!L1-D!5C`3L!Y_„!DX{!

cosSNH
!
=
21
7
⇒ tanSNH
!
=
2
3
⇒ SH = HN .tanSNH
!
=
a 3

2
*!!!

S
ABCD
= 4S
HAB
= 2HN .AB = 2.a 3.
3a
4
=
3a
2
3
2
*!
YCG!JE{!

V
S . ABCD
=
1
3
SH .S
ABCD
=
1
3
.
a 3

2
.
3a
2
3
2
=
3a
3
4
:75 ;*!
z;!=E!DX{!

HS = HA = HC =
a 3
2
!343!_!]&!.M'!7Oo3L!.Jt3!3L%?1!.12F!.E'!L1-D!YZ#*!
Ba.!V$-D![\!5C`3L!LXD!5I1!'a.!F$b3L!:YZ#;!343![\!]&!.JKD!7Oo3L!.Jt3!3L%?1!.12F!.E'!L1-D!
YZ#*!
Ny1!h!]&!.M'!.E'!L1-D!7}C!Z#\!.E!DX!

OA = OC = OD = OS = a
343!h!]&!.M'!'a.!DTC!3L%?1!.12F!
.U!g1W3!Y*Z#\!5&!0-3!VR3$!0e3L!E*!!!!
Câu%6%(1,0%điểm).!=J%3L!V$`3L!L1E3!5I1!$W!.JKD!.%?!7@!hiGj!D$%!71A'!Z:"klkm";!5&!'a.!F$b3L!

(P ) : 2x − 2y + z + 6 = 0
*!n12.!F$OP3L!.J>3$!7Oo3L!.$b3L!g!71!pCE!Z!5&!5C`3L!LXD!5I1!'a.!
F$b3L!:q;*!=>'!.%?!7@!71A'!B!.$%,!'r3!

AM = 2
5&!V$%,3L!D-D$!.s!B!723!:q;!7?.!L1-!.J9!]I3!
3$Q.*!
z;!|Oo3L!.$b3L!g!5C`3L!LXD!5I1!:q;!343!3$•3!5.F.!DHE!:q;!]&'!5…D!.P!D$u!F$OP3L^!(CG!
JE{

u
!
d
= (2;−2;1)
*!\%!7X{!

d :
x −1
2
=
y
−2
=
z +1
1
*!
z;!|A!V$%,3L!D-D$!.s!B!723!:q;!]I3!3$Q.!V$1!B!3e'!.J43!7Oo3L!.$b3L!g!5&!B!DX!V$%,3L!D-D$!
723!:q;!]I3!$P3!V$%,3L!D-D$!.s!Z!723!:q;^!V$1!7X!!

M (1+ 2t;−2t;−1+ t )
*!
=E!DX{!

AM = 2 ⇔ 4t

2
+ 4t
2
+ t
2
= 4 ⇔
t = −
2
3
t =
2
3
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⇒
M (
7
3
;−
4
3
;−
1
3

)
M (−
1
3
;
4
3
;−
5
3
)
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
*!
Kiểm%tra:%
z;!nI1!

M (
7
3
;−
4
3
;−

1
3
) ⇒ d(M ;(P )) =
13
3
*!
z;!nI1!

M (−
1
3
;
4
3
;−
5
3
) ⇒ d(M ;(P )) =
1
3
*!
Y%!(-3$!$E1!.JOo3L!$xF!.E!DX!71A'!DT3!.>'!]&!

M (
7
3
;−
4
3
;−

1
3
)
*!
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†!
Kết%luận:!n•G!

M (
7
3
;−
4
3
;−
1
3
)
*!!!!!!!
Cách%2:!Ny1!

M (a;b;c ) ⇒ AM
2
= 4 ⇔ (a −1)
2
+ b
2
+ (c +1)

2
= 4
*!
=E!DX{!
!

d (M ;(P )) =
2a −2b + c + 6
3
=
2(a −1)− 2b + (c +1)+ 7
3
≤
2(a −1)− 2b + (c +1) + 7
3
≤
(2
2
+ (−2)
2
+1
2
)((a − 1)
2
+ b
2
+ (c +1)
2
) + 7
3

=
13
3
*!
\QC!0e3L!i,G!JE!V$1!5&!D$u!V$1!

(a −1)
2
+ b
2
+ (c +1)
2
= 4
a −1
2
=
b
−2
=
c +1
1
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪

⎪
⎪
⇔ a =
7
3
,b = −
4
3
,c = −
1
3
*!
=E!DX!V2.!pC,!.OP3L!./*!!
Câu%7%(1,0%điểm).!=J%3L!'a.!F$b3L!5I1!.JKD!.%?!7@!hiG!D$%!.E'!L1-D!L1-D!Z[#!3@1!.12F!7Oo3L!
.Jt3!

(C ) : x
2
+ (y−
29
8
)
2
=
697
64
*!=>'!.%?!7@!7u3$!Z^!012.!7Oo3L!.Jt3!3L%?1!.12F!.E'!L1-D!v[#!Dw.!
Z[^Z#!]T3!]Ox.!.?1!D-D!71A'!

M (

21
10
;1), N (0;
13
5
),(M ≠ B, N ≠ C )
!5&!Z!DX!$%&3$!7@!M'*!
!
+)%Phát%hiện%tính%chất%hình%học:%%
=E!DX!Zv!5C`3L!LXD!5I1!B„^!.$•.!5•G{!
Ny1!_!]&!L1E%!71A'!DHE!Zv!5&!B„^!.E!DX{!
=$f%!.R3$!D$Q.!LXD!d!.M'!.E!DX{!
!

AIC
!
= 2ABC
!
⇒ IAC
!
= 90
0
− ABC
!
*!
z;!=U!L1-D![B„#!3@1!.12F!7Oo3L!.Jt3!343!

ANH
!
= ABC

!
*!
#@3L!]?1!.$f%!52!.E!7OxD{!
!

IAC
!
+ ACH
!
= 90
0
⇒ AHC
!
= 90
0
*!
n•G!Zv!5C`3L!LXD!5I1!B„*!!
z;!!
!
!
z;!|Oo3L!.$b3L!Zv!71!pCE!v!5&!5C`3L!LXD!5I1!B„!DX!F$OP3L!.J>3$!]&!

21x −16y + 58 = 0
*!
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21x −16y + 58 = 0
x
2
+ (y−

29
8
)
2
=
697
64
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇔
x = −2, y =1
x = 2, y =
25
4
⎡
⎣
⎢
⎢
⎢
⎢
⇒ A(−2;1);A(2;
25
4

)
*!
Kết%luận:!n•G!Z:m<k";!$%aD!Z:<k<†‡€;*!!!
Câu%8%(1,0%điểm).!N1,1!$W!F$OP3L!.J>3$!

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

(x, y ∈ !)
*!
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!

(x
2
− xy +1)( y
2
− xy +1) = 1 ⇔ (x − y)
2
(xy −1) = 0 ⇔
x = y
xy =1
⎡
⎣
⎢
⎢
*!
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Chi%tiết:%Mathlinks.vn!
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3+
1
x
2
+ x

2
− x +1 = x +
1
x
+1
*!
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!

3+
1
x
2
+ x
2
− x +1+ 2 (3+
1
x
2
)(x
2
− x +1) = x
2
+
1
x
2
+ 3+
2
x

+ 2x
⇔ 2 (3+
1
x
2
)(x
2
− x +1) = 3x +
2
x
−1
!*!
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!

2 (3+
1
x
2
)(x
2
− x +1) = (3+
1
x
2
)((2− x )
2
+ 3x
2
) ≥

2− x
x
+ 3x
⇒VT
(*)
≥VP
(*)
*!
n>!5•G!gQC!0e3L!F$,1!i,G!JE!

⇔
2− x
1
x
=
3x
3
> 0 ⇔ x = 1
*!
Kết%luận:!_W!q$OP3L!.J>3$!DX!3L$1W'!gCG!3$Q.!

(x; y) = (1;1)
*!!
Bình%luận:!=E!DX!.$A!0>3$!F$OP3L!$E1!52!7A!L1,1!F$OP3L!.J>3${!
!

2 (3+
1
x
2

)(x
2
− x +1) = 3x +
2
x
−1
*!
=$•G!5•G!0>3$!F$OP3L!$E1!52!.E!7OxD{

(x −1)(5x
3
+ 3x
2
+ 8) = 0 (**)
*!!
#$‰!‚!„2C!i•l!.$>!

x
2
− x +1 >1, x +
1
x
+1<1
F$OP3L!.J>3$!5`!3L$1W'*!
n>!5•G!!

(**) ⇔ x = 1
*!!!
Bài%tập%tương%tự%x!N1,1!$W!F$OP3L!.J>3$!

x
2
− xy +1 + y
2
− xy +1 = (x − y)
2
+ 4
1
x
2
+ 3 + x
2
− x +1 = y +
1
y
+1
⎧
⎨
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
*!
Câu%9%(1,0%điểm).%#$%!i^G^j!]&!D-D!()!F$UD!.$%,!'r3!

x = y = z = 1
*!=>'!L1-!.J9!]I3!3$Q.!DHE!
01AC!.$UD!

P = x + y − z + y + z − x + z + x − y
*!
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x
z
=
y
z
= 1
*!!
+$1!7X!

P = z .(
x
z
+
y
z
−1 +
y
z
+1−
x
z
+ 1+

x
z
−
y
z
) =
x
z
+
y
z
−1 +
y
z
+1−
x
z
+ 1+
x
z
−
y
z
*!
=E!7a.!

a =
x
z
,b =

y
z
⇒ P = a + b −1 + b +1− a + 1+ a− b , a = b =1
*!
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0 ≤ α ≤ β < 2π
(E%!D$%!

a = cosα + i.sinα,b = cosβ +i.sin β
*!
+$1!7X{!
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Chi%tiết:%Mathlinks.vn!
Ž!
!

a + b −1 = (cosα + cosβ −1)
2
+ (sinα + sin β)
2
= 3− 2cosα −2cosβ + 2cosα cosβ + 2sinαsin β
= 3− 4cos
α + β
2
.cos
α− β
2
+ 2cos(α− β)

= 1− 4cos
α + β
2
.cos
α− β
2
+ 4cos
2
α− β
2
= 1− 4cosu.cosv + 4cos
2
v
*!
nI1!

u =
α + β
2
,v =
α− β
2
*!
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!

b +1− a = 1− 4sinu.sinv + 4sin
2
v ,
a +1−b = 1+ 4sinu.sinv + 4sin

2
v
*!
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P = 1− 4cosu.cosv + 4cos
2
v + 1− 4sinu.sinv + 4sin
2
v + 1+ 4sinu.sinv + 4sin
2
v
≤ 3(7− 4 cosu.cosv + 4sin
2
v) ≤ 21+12cosv +12sin
2
v
= 33+12cosv −12cos
2
v = 36−12(cosv −
1
2
)
2
≤ 6
*!
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cosv =

1
2
cosu = −1
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇔
α + β = 2π
β −α =
2π
3
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⇔
α =
2π
3

β =
4π
3
⎧
⎨
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
*!
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z;!=E!DT3!D$‰!‚!$E1!D`3L!.$UD!]143!pCE3!723!'`7C3!DHE!()!F$UD!3$O!(EC{!
!

z
1
z
2
=
z

1
z
2
, z
1
.z
2
= z
1
. z
2
*!!
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‹YD$ŒEJ.j!g?3L{!
!

x
1
+ x
2
+ + x
n
≤ n(x
1
+ x
2
+ + x
n
)
*!!

z;!„2C!D-D!0&1!.%-3!G4C!DTC!5}!L1-!.J9!3$’!3$Q.!]‰D!7X!D$‰!‚!(Š!gK3L!0Q.!7b3L!.$UD!B13D%F(V1!
g?3L{!
!

x
1
2
+ y
1
2
+ x
2
2
+ y
2
2
+ + x
n
2
+ y
n
2
≥ (x
1
+ x
2
+ + x
n
)
2

+ (y
1
+ y
2
+ + y
n
)
2
!*!
\QC!0e3L!7?.!.?1!

x
1
y
1
=
x
2
y
2
= =
x
n
y
n
*!!
Bài%tập%tương%tựx%#$%!i^G^j!]&!D-D!()!F$UD!.$%,!'r3!

x = y = z = 1
*!=>'!L1-!.J9!]I3!3$Q.!DHE!

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Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%%
Chi%tiết:%Mathlinks.vn!
“!
01AC!.$UD!

P = x + y − z + y + z − x + z + x − y
*!