Số hóa bởi trung tâm học liệu />n n > 3, n ∈ N
Số hóa bởi trung tâm học liệu />Số hóa bởi trung tâm học liệu />Số hóa bởi trung tâm học liệu />Số hóa bởi trung tâm học liệu />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.
P
n
(x) P
n
. deg P
n
(x) = n
a
n
a
0
a
n
x
n
P
n
(x) = a
n
x
n
+ a
n−1
x
n−1
+ ··· + a
1
x + a
0
, a
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 />P
n
(z) = 0
a ∈ C P
n
(z) = 0 P
n
(a) = 0
0 = P
n
(a) = P
n
(a).
P
n
(x) = a
0
(x − α
1
)
n
1
. . . (x − α
r
)
n
r
(x
2
+ p
1
x + q
1
)
m
1
. . . (x
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
1
, . . . , α
r
; p
1
, q
1
, . . . p
s
, q
s
∈ R.
P
n
(x) n k k ≤ n
n k
n n
f(z) = a
0
z
n
+ a
1
z
n−1
+ ··· + a
n−1
z + a
n
(a
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
+ 4z
5
+ 3z
4
+ 4z
3
− 6z
2
+ z + 2.
Số hóa bởi trung tâm học liệu />f(z) n
z
n
f
1
z
= f(z), z = 0.
P (x) = a
n
x
n
+a
n−1
x
n−1
+a
n−2
x
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)
P (x) = 0 ⇔ (x − 1)Q (x) = 0,
deg Q (x) = 2k, Q(x)
2k
y = x +
1
x
k
x
4
+ 2x
3
− 6x
2
+ 2x + 1 = 0.
x
4
+ 2x
3
− 6x
2
+ 2x + 1 = 0. (1)
x = 0
(1) ⇔ x
2
+
2
x
2
+ 2(x +
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)
Số hóa bởi trung tâm học liệu />y = x +
1
x
⇔ y
2
+ 2y −8 = 0.
⇔
y = 2
y = −4
⇔
x +
1
x
= 2
x +
1
x
= −4
⇔
x
2
− 2x + 1 = 0
x
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
x
4
− 5x
3
+ 8x
2
− 5x + 1 = 0.
x
4
− 5x
3
+ 8x
2
− 5x + 1 = 0. (2)
x = 0
(2) ⇔
x
2
+
1
x
2
−5
x +
1
x
+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 +
1
x
= 3 ⇔ x
2
− 3x + 1 = 0 ⇔ x =
3 ±
√
5
2
x = 1; x = −1; x =
3 +
√
5
2
; x =
3 −
√
5
2
.
Số hóa bởi trung tâm học liệu />x
4
+ 2x
3
+ 4x
2
+ 2x + 1 = 0. (1)
x
4
+ 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 +
1
x
= |x| +
1
2
≥ 2,
(2) ⇔ y
2
+ 2y + 2 = 0. (3)
∆
= 1 − 2 = −1 < 0.
(1) ⇔ x
2
x
2
+ 2x + 1
+
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
2
+ ax + 1 = 0 a
2
+ b
2
x
4
+ ax
3
+ bx
2
+ ax + 1 = 0.(1)
x
0
= 0,
x
2
0
+
1
x
2
0
+ a
x
0
+
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 />y = x
0
+
1
x
0
y
2
0
+ ay
0
+ b − 2 = 0
⇔
2 − y
2
0
= |ay
0
+ b|. (3)
|ay
0
+ b| ≤
√
a
2
+ b
2
y
2
0
+ 1
⇔
2 − y
2
0
≤
√
a
2
+ b
2
y
2
0
+ 1
⇔ a
2
+ b
2
≥
2 − y
2
0
2
1 + y
2
0
≥
4
5
.
(4) ⇔ 5
2 − y
2
0
2
≥ 4
1 + y
2
0
(4)
⇔ 5y
4
0
− 24y
2
0
+ 16 ≥ 0 (5)
⇔
y
2
0
≥ 4.
y
2
0
≤
4
5
.
|y| =
x
0
+
1
x
0
y
0
= 2; a = −
4
5
; b = −
2
5
a
2
+ b
2
4
5
.
x = (x
1
, x
2
, . . . , x
n
) ∈ R
n
f(x) = f(x
1
, x
2
, . . . , x
n
)
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
j
n
x
j
1
1
x
j
2
2
. . . x
j
n
n
,
j
i
∈ N(i = 0, 1, 2 . . . n)
f(x) = (x
1
, x
2
, . . . , x
n
)
Số hóa bởi trung tâm học liệu />f(x) = (x
1
, x
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
, k ∈ Z,
σ
0
= 1,
σ
1
(x) = x
1
+ x
2
+ ··· + x
n
=
n
i=1
x
i
,
σ
2
(x) =
1≤i<j≤n
x
i
x
j
,
σ
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
1
+ x
2
+ x
3
+ x
4
,
σ
2
= x
1
x
2
+ x
1
x
3
+ x
1
x
4
+ x
2
x
3
+ x
2
x
4
+ 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
4
,
σ
4
= x
1
x
2
x
3
x
4
.
σ
k
C
k
n
=
n!
k!(n − k)!
x
1
, x
2
, . . . x
n
n
(x) y
1
, 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
2
2
. . . x
k
n
n
)
Số hóa bởi trung tâm học liệu />S
k
= σ
1
S
k−1
− σ
2
S
k−2
+ ··· + (−1)
i−1
σ
i
S
n−i
+ ··· + (−1)
k−1
kσ
k
. (1.1)
(−1)
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
+ m
2
+ ··· + m
k
− 1)!
m
1
!m
2
! . . . m
k
!
σ
m
1
1
σ
m
2
2
. . . σ
m
k
k
(1.2)
S
k
σ
j
f(x
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
i=1
x
i
,
σ
2
(x) =
1≤i<j≤n
x
i
x
j
,
σ
3
(x) =
1≤i<j<l≤n
x
i
x
j
x
l
,
. . .
σ
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
k
y
l
z
m
,
Số hóa bởi trung tâm học liệu />, k, l, m ∈ N x, y, z,
a
klm
∈ R
∗
= R\{0} ,
k + j + m ϕ(x, y, z).
P (x, y, z)
P (x, y, z) =
k+l+m≤n
a
klm
x
k
y
l
z
m
.
P (x, y, z)
P (x, y, z) = P (y, x, z) = P (x, z, y).
xy + yz + zx,
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
3
= xyz,
x, y, z
σ
1
= x + y + z, σ
2
= xy + yz + zx, σ
3
= xyz.
ϕ(t, u, v), ψ(t, u, v)
t = σ
1
= x + y + z, u = σ
2
= xy + yz + zx, v = σ
3
= xyz,
Số hóa bởi trung tâm học liệu />f
m
(x, y, z)
f
m
(x, y, z)
f
m
(x, y, z) =
i+2j+3k=m
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
,
f
3
(x, y, z) = a
1
σ
3
1
+ a
2
σ
1
σ
2
+ a
3
σ
3
,
f
4
(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
+ a
3
σ
1
σ
2
2
+ a
4
σ
2
1
σ
3
+ a
5
σ
2
σ
3
,
f
6
(x, y, z) = a
1
σ
6
1
+ a
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
+ a
7
σ
1
σ
2
σ
3
.
∆(x, y, z) = (x − y)
2
(x − z)
2
(y −z)
2
.
∆(x, y, z)
∆(x, y, z) = a
1
σ
6
1
+ a
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
+ a
7
σ
1
σ
2
σ
3
∆(x, y, z) a
1
=
a
2
= 0
a
4
= −4, a
3
= 1, a
6
= −27, a
5
= −4, a
7
= 18.
∆(x, y, z) = σ
2
1
σ
2
2
− 4σ
3
2
− 4σ
3
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
ax
k
y
l
= k + l.
Số hóa bởi trung tâm học liệu />k, l x, y
3x
2
y,
2
3
x
2
y
3
x, y
x, y
ax
k
y
l
, Bx
k
y
l
(A = B)
ax
k
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
5
P (x, y)
x, y
P (x, y) x, y
P (x, y) =
k+l≤m
a
kl
x
k
y
l
.
P (x, y)
P (x, y) = P (y, x).
P (x, y) = x
2
+ xy + y
2
, Q(x, y) = x
2
y + xy
2
σ
0
= 1, σ
1
= x + y, σ
2
= 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 />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
y
k
= a(xy)
k
= aσ
k
2
.
bx
k
y
l
(k = l)
bx
l
y
k
a k
b(x
k
y
l
+ x
l
y
k
) = bx
k
y
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
k
y
k
σ
1
, σ
2
ϕ(σ
1
, σ
2
), ϕ(σ
1
, σ
2
)
σ
1
= x + y, σ
2
= x.y P (x, y)
ϕ(σ
1
, σ
2
) ≡ ψ(σ
1
, σ
2
)
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
2
+ b
2
= S + 3
Số hóa bởi trung tâm học liệu />⇔
S + 3 − 3(a + b) = 3
(a + b)
2
− 2ab = S + 3
⇔
a + b =
S
3
ab =
S
2
− 9S − 27
18
a; b ≥ 0.
⇔
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) =
9 + 3
√
21
2
x = 0, y = 0
(x + y)xy = x
2
+ y
2
− xy.
A =
1
x
3
+
1
y
3
(x + y)xy = x
2
+ y
2
− xy ⇔
1
y
+
1
x
=
1
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 =
a −
b
2
2
+
3b
2
4
⇒ a + b ≥ 0
a + b = a
2
+ b
2
− ab
a + b = S
⇔
a + b = (a + b)
2
− 3ab
a + b = S
⇔
a + b = S
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 />S
k
= x
k
+ y
k
, (k = 1, 2 . . .)
k
S
m
= x
m
+y
m
σ
1
= x + y σ
2
= xy
s
1
= σ
1
,
s
2
= σ
2
1
− 2σ
2
,
s
3
= σ
3
1
− 3σ
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
σ
2
2
,
. . .
S
k
σ
1
, σ
2
S
k
k
=
[k/2]
m=0
(−1)
m
(k −m − 1)!
m!(k −2m)
σ
k−2m
1
σ
m
2
.
[k/2] k/2
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 S
n
= x
n
1
+ x
n
2
.
aS
n+2
+ bS
n+1
+ cS
n
= 0.
A = (1 +
√
2)
5
+ (1 −
√
2)
5
.
Số hóa bởi trung tâm học liệu />x
n+2
1
+ x
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
2
) − S
n
x
1
x
2
.
x
1
+ x
2
= −
b
a
x
1
x
2
=
c
a
S
n+2
= −
b
a
S
n+1
−
c
a
S
n
aS
n+2
+ bS
n+1
+ cS
n
= 0.
Số hóa bởi trung tâm học liệu />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
n
= 0
x
1
+ x
2
+ ··· + x
n
= σ
1
x
1
x
2
+ x
1
x
3
+ ··· + x
n−1
x
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
u
1
, u
2
, ··· , u
n
x
1
= u
1
, x
2
= u
2
, ··· , x
n
= u
n
x
1
, x
2
, ··· , x
n
u
1
, u
2
, ··· , 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 />
u
1
+ u
2
+ ··· + u
n
= σ
1
u
1
u
2
+ u
1
u
3
+ ··· + u
n−1
u
n
= σ
2
···
u
1
u
2
···u
n
= σ
n
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
x
1
, x
2
, ··· , x
n
u
1
, u
2
, ··· , u
n
x
1
= u
1
, x
2
= u
2
, ··· , x
n
= u
n
u
1
, u
2
, ··· , u
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
f(x)
α
1
+ α
2
+ ··· + 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
k
+ ··· + α
n−k+1
α
n−k+2
. . . α
n
= (−1)
k
a
k
a
0
. . . . . . . . . . . . . . . . . . . . .
α
1
α
2
. . . α
n
= (−1)
n
a
n
a
0
f(x) = a
0
(x−α
1
)(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 />··· + α
n−1
α
n
)x
n−2
+ ··· + (−1)
n
α
1
α
2
. . . α
n
].
α
1
+ α
2
+ ··· + 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
k
+ ··· + α
n−k+1
α
n−k+2
. . . α
n
= (−1)
k
a
k
a
0
. . . . . . . . . . . . . . . . . . . . .
α
1
α
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
, ··· , x
n
σ
1
, σ
2
, ··· , σ
n
σ
1
= −a
n−1
, σ
2
= a
n−2
.
x
2
1
+ x
2
2
+ ···+ x
2
n
= s
2
= σ
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
y + 1 = 0.
y
2
1
+ y
2
2
+ ··· + y
2
n
= 3.
9 = (x
2
1
+ x
2
2
+ ··· + x
2
n
)
1
x
2
1
+
1
x
2
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
+ t
3
= 1
x
4
+ y
4
+ z
4
+ t
4
= 33
Số hóa bởi trung tâm học liệu />x + y + z + t = σ
1
, xy + xz + xt + yz + yt + zt = σ
2
,
xyz + xyt + xzt + yzt = σ
3
, xyzt = σ
4
σ
1
= 1
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
3
− 4σ
4
= 33
σ
1
= 1, σ
2
= −4, σ
3
= 4, σ
4
= 0.
u
4
− u
3
− 4u
2
+ 4u = 0 ⇔ u(u − 1)(u − 2)(u + 2) = 0.
u
1
= 0, u
2
= 1, u
3
= 2, u
4
= −2.
(x, y, z, t) = (0, 1, 2, −2)
x
1
+ x
2
+ ··· + x
n
= a
x
2
1
+ x
2
2
+ ··· + x
2
n
= a
2
···
x
n
1
+ x
n
2
+ ··· + x
n
n
= a
n
σ
1
, σ
2
, ··· , σ
n
x
1
, x
2
, ··· , x
n
s
k
(k = 1, 2, ··· , n) x
k
1
+x
k
2
+···+x
k
n
=
s
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
(k = 2, 3, . . . , n). x
1
, x
2
, ··· , x
n
x
1
, x
2
, ··· , x
n
x
i
= a, x
j
= 0 : j = i = 1, 2, ··· ,
(a, 0, . . . , 0), (0, a, 0, . . . , 0), . . . (0, 0, . . . , a).
Số hóa bởi trung tâm học liệu />P (x, y, z), Q(x, y, z), R(x, y, z)
P (x, y, z) = 0
Q(x, y, z) = 0
R(x, y, z) = 0
x + y + z = σ
1
, xy + yz + zx = σ
2
, 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
2
, σ
3
σ
1
, σ
2
, σ
3
x, y, z
σ
1
, σ
2
, σ
3
u
3
− σ
1
u
2
+ σ
2
u − σ
3
= 0
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
1
y
2
= u
3
z
2
= u
2
;
x
3
= u
2
y
3
= u
1
z
3
= u
3
x
4
= u
2
y
4
= 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
z
6
= u
2
x = a, y = b, z = c
a, b, c
u
1
, u
2
, u
3
u
3
− σ
1
u
2
+ σ
2
u − σ
3
= (u − u
1
)(u − u
2
)(u − u
3
).
Số hóa bởi trung tâm học liệu />