Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
n n > 3, n ∈ N
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
n x
P
n
(x) = a
n
x
n
+ a
n−1
x
n−1
+ ··· + a
1
x + a
0
,
a
n
, a
n−1
, . . . , a
0
a
n
= 0, n ∈ N.

n
= 0.
α ∈ C P
n
(x) P
n
(α) = 0.
k ∈ N, k > 1 P
n
(x) (x − α)
k
P
n
(x)  (x − α)
k+1
α
k P
n
(x)
k = 1 α k = 2 α
n ≥ 1
C n
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
P
n
(z) = 0
a ∈ C P
n
(z) = 0 P
n

2
s
+ p
s
x + q
s
)
m
s
,
r

i=1
n
i
+2
s

i=1
m
i
= n, p
2
i
− 4q
i
< 0, i = 1, s
α
0
, α

0
= 0)
a
0
= a
n
, a
1
= a
n−1
, a
2
= a
n−2
, . . .
z
5
− 3z
4
+ 2z
3
+ 2z
2
− 3z + 1,
2z
8
+ z
7
− 6z
6

n−2
+···+a
1
x
2
+a
1
x+a
0
= 0, a
n
= 0 (1)
a
n
= a
0
; a
n−1
= a
1
; a
n−2
= a
2
; . . .
n = 2k + 1,
n = 2k
x = −1
P (x) = 0
(deg P (x) = 2k + 1)

x
) − 6 = 0.
⇔

x +
1
x

2
− 2 + 2

x +
1
x

− 6 = 0.
⇔

x +
1
x

2
+ 2

x +
1
x

− 8 = 0. (2)

+ 4x + 1 = 0
⇔

x = 1
x = −2 ±
√
3.
x = 1; x = −2 +
√
3; x = −2 −
√
3.
x
5
− 4x
4
+ 3x
3
+ 3x
2
− 4x + 1 = 0. (1)
(x + 1) (x
4
− 5x
3
+ 8x
2
− 5x + 1) = 0 ⇔

x = −1

+8 = 0 ⇔

x +
1
x

2
−5

x +
1
x

+6 = 0. (3)
y = x +
1
x
y
2
− 5y + 6 = 0 ⇔

y = 2.
y = 3.
x +
1
x
= 2 ⇔ x
2
− 2x + 1 = 0 ⇔ x = 1.
x +

+ 2x
3
+ 4x
2
+ 2x + 1 = 0. x = 0
(1) ⇔

x
2
+
1
x
2

+ 2

x +
1
x

+ 4 = 0. (2)
y = x +
1
x
|y| =




x +


+

x
2
+ 2x + 1

+ 2x
2
= 0.
⇔

x
2
+ 1

(x + 1)
2
+ 2x
2
= 0.
⇔

x + 1 = 0. (4)
x = 0. (5)
x
4
+ ax
3
+ bx

1
x
0

+ b = 0
⇔

x
0
+
1
x
0

2
+ a

x
0
+
1
x
0

+ b − 2 = 0. (2)
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
y = x
0
+
1

0
+ 1
⇔


2 − y
2
0


≤
√
a
2
+ b
2

y
2
0
+ 1
⇔ a
2
+ b
2
≥

2 − y
2
0

⇔

y
2
0
≥ 4.
y
2
0
≤
4
5
.
|y| =




x
0
+
1
x
0




y
0

f(x) =
m

k=0
M
k
(x)
M
k
(x) = M
k
(x
1
, x
2
, . . . , x
n
) =

j
1
+j
2
+···+j
n
=k
a
j
1
j

2
, . . . , x
n
)
f(tx
1
, tx
2
, . . . , tx
n
) = t
m
f(x
1
, x
2
, . . . , x
n
), ∀t = 0.
S
k
= x
k
1
+ x
k
2
+ ··· + x
k
n

3
(x) =

1≤i<j<l≤n
x
i
x
j
x
l
,
. . .
σ
n
(x) = x
1
x
2
. . . x
n
.
S
k
σ
r
(x), (r = 1, 2, 3, . . . , n)
n
σ
1
= x

+ x
3
x
4
,
σ
3
= x
1
x
2
x
3
+ x
1
x
2
x
4
+ x
2
x
3
x
4
+ x
1
x
3
x

, y
2
, . . . , y
n
n
(y)
(x) (y) (x) ∼ (y)
λ ∈ R, λ = 0, x
i
= λy
i
i = 1, 2, . . . , n.
x
k
1
1
x
k
2
2
. . . x
k
n
n
o(x
k
1
1
x
k

i−1
σ
i
S
n−i
= 0 i > n
S
k
S
k
k
=

m
1
+2m
2
+···+km
k
=k
(−1)
k−m
1
−m
2
−···−m
k
.
(m
1

1
, x
2
, . . . , x
n
)
n Φ(σ
1
, σ
2
, σ
3
. . . σ
n
)
σ
1
,σ
2
, . . . ,σ
n
σ
1
(x) = x
1
+ x
2
+ ··· + x
n
=

n
(x) = x
1
x
2
. . . x
n
.
f(x) = f(x
1
, x
2
, . . . , x
n
) σ
1
,σ
2
, . . . ,σ
n
x
1
, x
2
, . . . , x
n
ϕ(x, y, z)
ϕ(x, y, z) = a
klm
x

x
3
+ y
3
+ z
3
− 3xyz,
(x + y)(y + z)(z + x),
x(y
4
+ z
4
) + y(x
4
+ z
4
) + z(y
4
+ x
4
).
f(x, y, z)
f(tx, ty, tz) = t
m
f(x, y, z), t = 0.
σ
1
= x + y + z, σ
2
= xy + yz + zx, σ

a
ijk
σ
i
1
σ
j
2
σ
k
3
(j, i, k ∈ N).
f
1
(x, y, z) = a
1
σ
1
,
f
2
(x, y, z) = a
1
σ
2
1
+ a
2
σ
2

1
σ
2
+ a
3
σ
2
2
+ a
4
σ
2
1
σ
3
,
f
5
(x, y, z) = a
1
σ
5
1
+ a
2
σ
3
1
σ
2

σ
4
1
σ
2
+ a
3
σ
2
1
σ
2
2
+ a
4
σ
3
2
a
5
σ
3
1
σ
3
+ a
6
σ
2
3

3
σ
2
1
σ
2
2
+ a
4
σ
3
2
+ a
5
σ
3
1
σ
3
+ a
6
σ
2
3
+ a
7
σ
1
σ
2

1
σ
3
− 27σ
2
3
+ 18σ
1
σ
2
σ
3
.
f(x, y)
x, y
f(x, y) = a
kl
x
k
y
l
,
a
kl
= 0 k, i
a
kl
k + l f(x, y)
deg [f(x, y)] = deg


y
l
, Bx
m
y
n
x, y ax
k
y
l
Bx
m
y
n
x, y k > m k = m, l > n
x
4
y
2
x
2
y
7
a x
4
y
6
x
4
y

= xy
σ
j
(j = 0, 1, 2)
f(x, y)
f(tx, ty) = t
m
f(x, y), ∀t = 0.
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
P (x, y)
p(σ
1
, σ
2
)
σ
1
= x + y, σ
2
= xy
P (x, y) = p(σ
1
, σ
2
).
ax
k
y
k
ax

k
(x
l−k
+ y
l−k
) = bσ
k
2
s
l−k
.
s
l−k
σ
1
, σ
2
σ
1
, σ
2
b(x
k
y
l
+ x
l
y
k
), ax

x, y
x − 3
√
x + 1 = 3

y + 2 − y.
x + y.
x + y = S.

x − 3
√
x + 1 = 3
√
y + 2 − y.
x + y = S.
√
x + 1 = a;
√
y + 2 = b a; b ≥ 0 x = a
2
− 1; y = b
2
− 2

a
2
+ b
2
− 3(a + b) = 3
a






S
3

2
≥ 4.
S
2
− 9S − 27
18
.
S ≥ 0.
S
2
− 9S − 27 ≥ 0.
⇔
9 + 3
√
21
2
≤ S ≤ 9 + 3
√
15.
max(x + y) = 9 + 3
√
15; min(x + y) =

y
2
+
1
x
2
−
1
xy
.
1
x
= a;
1
y
= b
a + b = a
2
+ b
2
− ab A = a
3
+ b
3
.
a + b = a
2
+ b
2
−ab =

S
2
− S
3
⇔ S
2
≥
4(S
2
− S)
3
⇔ S
2
− 4S ≤ 0 ⇔ 0 ≤ S ≤ 4
⇒ (a + b)
2
≤ 16.
a = b = 2 ⇔ x = y =
1
2
max A = 16.
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
S
k
= x
k
+ y
k
, (k = 1, 2 . . .)
k

1
σ
2
,
s
4
= σ
4
1
− 4σ
2
1
σ
2
+ 2σ
2
2
,
s
5
= σ
5
1
− 5σ
3
1
σ
2
+ 5σ
1

S
k
= x
k
+ y
k
+ z
k
k σ
1
, σ
2
, σ
3
.
x, y, z
σ
1
= x+y+z, σ
2
= xy +yz +zx, σ
3
= xyz.
x
1
, x
2
ax
2
+ bx + c = 0, (a = 0).

n+2
2
= (x
n+1
1
+ x
n+1
2
)(x
1
+ x
2
) − (x
n
1
+ x
n
2
)x
1
x
2
.
S
n+2
= S
n+1
(x
1
+ x

c
a
S
n
aS
n+2
+ bS
n+1
+ cS
n
= 0.
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
n n > 3, n ∈ N
σ
1
, σ
2
, ··· , σ
n
u
n
− σ
1
u
n−1
+ σ
2
u
n−2
− ··· + (−1)

n
= σ
2
···
x
1
x
2
···x
n
= σ
n
u
1
, u
2
, ··· , u
n
n! x
1
= u
1
, x
2
= u
2
, ··· , x
n
= u
n

f(u) = u
n
−σ
1
u
n−1
+σ
2
u
n−2
−···+ (−1)
n
σ
n
= (u −u
1
)(u−u
2
) ···(u−u
n
).
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/







u

x
1
= u
1
, x
2
= u
2
, ··· , x
n
= u
n
u
1
, u
2
, ··· , u
n
x
1
, x
2
, ··· , x
n
σ
1
, σ
2
, ··· , σ
n

f(x) = a
0
x
n
+ a
1
x
n−1
+ ··· + a
n−1
x + a
n
∈ A[x], a
0
= 0
f(x) = a
0
(x − α
1
)(x − α
2
) . . . (x − α
n
)
α
1
, α
2
, . . . , α
n

+ ··· + α
n
= −
a
1
a
0
α
1
α
2
+ α
2
α
3
+ ··· + α
n−1
α
n
=
a
2
a
0
. . . . . . . . . . . . . . . . . . . . .
α
1
α
2
. . . α

)(x−α
2
) . . . (x−α
n
)
f(x) = a
0
[x
n
− (α
1
+ α
2
+ + α
n
) + (α
1
α
2
+ ··· + α
1
α
n
+ α
2
α
3
+
Số hóa bởi trung tâm học liệu http://www.lrc-tnu.edu.vn/
··· + α










α
1
+ α
2
+ ··· + α
n
= −
a
1
a
0
α
1
α
2
+ α
2
α
3
+ ··· + α
n−1

2
. . . α
n
= (−1)
n
a
n
a
0
x
n
+ a
n−1
x
n−1
+ a
n−2
u
n−2
+ ···+ a
1
x +a
0
= 0 (a
i
= ±1, i = 0, 1, ··· , n −1).
n n ≥ 3
n x
1
, x

2
1
−2σ
2
= a
2
n−1
−2a
n−2
= 3 ( |a
i
= 1|, s
2
≥ 0)
y =
1
x
i
y
i
a
0
y
n
+ a
1
y
n−1
+ ··· + a
n−1

2
+ ··· +
1
x
2
n

≥ 3
n ≥ 3.





x + y + z + t = 1
x
2
+ y
2
+ z
2
+ t
2
= 9
x
3
+ y
3
+ z
3

σ
2
1
− 2σ
2
= 9
σ
3
1
− 3σ
1
σ
2
+ 3σ
3
= 1
σ
4
1
− 4σ
2
1
σ
2
+ 2σ
2
2
+ 4σ
1
σ






x
1
+ x
2
+ ··· + x
n
= a
x
2
1
+ x
2
2
+ ··· + x
2
n
= a
2
···
x
n
1
+ x
n
2

k

σ
1
= a
s
2
= σ
2
1
− 2σ
2
= a
2
σ
2
= 0, s
1
= a = σ
1
, s
2
= a
2
= σ
2
1
.
σ
k

, xyz = σ
3
.



p(σ
1
, σ
2
, σ
3
) = 0
q(σ
1
, σ
2
, σ
3
) = 0
r(σ
1
, σ
2
, σ
3
) = 0
σ
1
, σ

x + y + z = σ
1
xy + yz + zx = σ
2
xyz = σ
3
u
1
, u
2
, u
3

x
1
= u
1
y
1
= u
2
z
1
= u
3
;

x
2
= u

= u
3
z
4
= u
1
;

x
5
= u
3
y
5
= u
2
z
5
= u
1
;

x
6
= u
3
y
6
= u
1