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7/23/2019 Fractions Ration Nelles Des
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K R C
P Q
P Q
K[X ]
Q = 0
Q K(X )
K
R
P Q
P
Q
F = P Q
(P )−
(Q)
F
(F )
(X2−2X+4)(X−2)X3
(X−
2)
2
(X2−2X+4)X3(X−2) 3 − 5 = −2
−∞
K(X ) +
×
P
Q +
R
S =
P S + QR
QS
P
Q ×
R
S =
P R
QS ,
X K
F = P Q
x → F (x) = P (x)Q(x)
F
K
Q
F
F = P Q ∈ K(X ) α ∈ K
α
F
α
P
α
F
α
Q
R(X )
(X 2 + X + 1)(X − 1)2X
(X − 2)(X 2 + 1)(X + 1)4
4
R(X )
F = P Q ∈ K(X )
E
G
F = E + G
(G) < 0.
E F P
Q
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F = P Q
•
(F ) < 0
•
(F ) ≥ 0
P
Q
P = QE +R (R) < (Q)
P
Q = QE +R
Q =
QE
Q +
R
Q = E
+ R
Q
<0
.
F 0 = X
X 2 − 4 0
F 1 = X 5 + 1
X (X − 1)2 X 2 + 2X + 3
F 2 = 1
(X 2 − 1)(X 2 + 1)2
0
F 3 = 4X 3
(X 2 − 1)2 0
R
F = P Q ∈ R(X )
E
Q
R[X ]
Q = λ
rk=1
(X − αk)mk
sl=1
(X 2 + β lX + γ l)nl .
(Ak,i) 1≤k≤r1≤i≤mk
(Bl,j) 1≤l≤s1≤j≤nl
(C l,j) 1≤l≤s1≤j≤nl
F = E
+r
k=1
mki=1
Ak,i
(X − αk)i
αk
+sl=1
nlj=1
Bl,jX + C l,j(X 2 + β lX + γ l)j
.
F R
F = P Q ∈ R(X )
m
α
Q = (X − α)m
Q1
Q1 ∈R
[X ]
Q1(α) = 0
F
F = A1
(X − α) +
A2
(X − α)2 + · · · +
Am
(X − α)m + F 0 =
mi=1
Ai
(X − α)i
α
+ F 0,
F 0
α
Ai
i = 1, . . . , m
A,B,C,D...
F 0 = X
X 2 − 4 F 0 =
A
(X − 2)
2
+ B
(X + 2)
−2
.
F 1 = X 5 + 1
X (X − 1)2 F 1 = X 2 + 2X + 3
+ A
X
0
+ B
X − 1 +
C
(X − 1)2
−1
.
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F 2 = 1
(X 2 − 1)(X 2 + 1)2
F 2 = A
X − 1 +
B
X + 1 +
C X + D
X 2 + 1 +
EX + F
(X 2 + 1)2.
F 3 = 4X 3
(X 2 − 1)2 F 3 =
A
X − 1 +
B
(X − 1)2 +
C
X + 1 +
D
(X + 1)2
R(X )
α
m
F
1(X−α)m F F (X − α)m
X
α
F − A(X−α)m
A
α
m− 1
F 0 = X
X 2 − 4 F 0 = A
(X−2) + B(X+2)
•
A
α = 2
m = 1
(X − 2)
X
X 2 − 4 × (X − 2)
X=2
=
A
(X − 2) × (X − 2) +
B
(X + 2) × (X − 2)
X=2
⇐⇒ X
X + 2
X=2
= 1
2 = A.
•
B
(X + 2)
−2
B = 12
F 0 F 0 = 12(X−2)
+ 12(X+2)
F 1 = X 5
+ 1X (X − 1)2
F 1 = X 2 + 2X + 3 + AX
+ BX−1 + C
(X−1)2
•
A
C
X 0 A = 1
(X − 1)2
1
C = 2
F 1 = X 2 + 2X + 3 + 1X
+ BX−1 + 2
(X−1)2
•
B
F 1 − 2
(X − 1)2 =
X 5 + 1
X (X − 1)2 −
2
(X − 1)2 =
X 5 − 2X + 1
(X − 1)2X .
X 5 − 2X + 1
X − 1
X 5− 2X + 1 = (X − 1)(X 4 + X 3 + X 2 + X − 1)
F 1 −
2
(X − 1)2 =
X 4 + X 3 + X 2 + X − 1
X (X − 1) .
1
B = 3 F 1 F 1 = X 2 + 2X + 3 + 1X
+ 3X−1 + 2
(X−1)2
X
F 1 = X 5 + 1
X (X − 1)2
F 1 = X 2 + 2X + 3 + 1X
+ BX−1
+ 2(X−1)2
X
F 1(−1) = (−1)5 + 1
(−1)(−1 − 1)2 = (−1)2 + 2(−1) + 3 +
1
−1 +
B
−1− 1 +
2
(−1 − 1)2 ⇐⇒ 0 =
−B
2 +
3
2,
B = 3
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F α m F
−α
m
F
F (X )
F (−X ) = ±F (X )
F
F 2 = 1
(X 2 − 1)(X 2 + 1)2
F 2(X ) = F 2(−X )
F 2(X ) = A
X − 1 +
B
X + 1 +
CX + D
X 2 + 1 +
EX + F
(X 2 + 1)2 =
−A
X + 1 + −B
X − 1 +−CX + D
X 2 + 1 +
−EX + F
(X 2 + 1)2 = F 2(−X ).
A = −B C = E = 0
F 2(X ) = A
X − 1 −
A
X + 1 +
D
X 2 + 1 +
F
(X 2 + 1)2.
•
A
(X − 1)
X = 1
A = 1/8
•
F
(X 2
+ 1)2
X = i
F = −1/2
•
D
0
X
−1 = −1/8− 1/8 + D − 1/2
D = −1/4
F 2(X ) = 18(X−1)
− 18(X+1)
− 14(X2+1)
− 12(X2+1)2 .
F x →xF (x)
F
F 3 = 4X 3
(X 2
− 1)2
F 3 = A
X − 1
+ B
(X − 1)2
+ C
X + 1
+ D
(X + 1)2
•
F 3
−F 3(X ) = −A
X − 1 +
−B
(X − 1)2 + −C
X + 1 +
−D
(X + 1)2 =
−A
X + 1 +
B
(X + 1)2 + −C
X − 1 +
D
(X − 1)2 = F 3(−X ).
A = C B = −D F 3 = AX−1 + B
(X−1)2 + AX+1 + B
(X+1)2
• B (X − 1)2 1 B = 1
• A limx→∞ xF 3(x) = limx→∞
4x4
(x2−1)2 = 4
limx→∞
xF 3(x) = limx→∞
Ax
x− 1 +
x
(x− 1)2 +
Ax
x + 1 +
x
(x + 1)2 = 2A.
2A = 4
A = 2
F 3 = 2
X − 1 +
1
(X − 1)2 +
2
X + 1 +
1
(X + 1)2.
F = X 2 + 3X + 1
(X − 1)2(X − 2), G =
X 5
X 4 − 1 H =
4
(X 2 − 1)2
F = 1
X(X+1)
G = 1X3(X3+1)
(a3 + b3) = (a + b)(a2 − ab + b2) a, b ∈ R.
