ułamki proste kajka: rozłóż na ułamki proste

	4x2−2x+1
f(x)=
	x4−x3−x+1

proszę chociaż o napisanie wyniku

Adamm: http://www.wolframalpha.com/input/?i=(4x%5E2-2x%2B1)%2F(x%5E4-x%5E3-x%2B1) "Partial fraction expansion"

Mila: W(x)=x⁴−x³−x+1=x³*(x−1)−(x−1)=(x−1)*(x³−1)=(x−1)*(x−1)*(x²+x+1)=(x−1)²*(x²+x+1)

4x2−2x+1		A		B		Cx+D
	=		+		+		=
(x−1)2*(x2+x+1)		x−1		(x−1)2		x2+x+1)

	A*(x3−1)+B*(x2+x+1)+(Cx+D)*(x−1)2
=		⇔
	(x−1)2*(x2+x+1)

4x²−2x+1=A*(x³−1)+B*(x²+x+1)+(Cx+D)*(x−1)² 1) x=0 L=1, P=A*(−1)+B*1+D*1⇔−A+B+D=1 2) x=1 L=4−2+1=3, P=A*0+B*3+(C+D)*0=3B 3B=3⇔B=1⇒−A+D=0⇔A=D 3) x=−1 L=4+2+1=7 , P=A*(−2)+B*(1−1+1)+(C*(−1)+D)*(−1−1)²=−2A+B+(−C+D)*4, P=−2A+B−4C+4D=−2A+1−4C+4A=2A−4C+1 2A−4C+1=7, 2A−4C=6 4) x=2 L=16−4+1=13 P=A*(8−1)+B*(4+2+1)+(C*2+D)*(2−1)²=7A+7B+2C+D=7A+7+2C+A=8A+2C+7 8A+2C+7=13⇔8A+2C=6 4A+C=3 i 2A−4C=6 stąd : A=1, C=−1, ======== A=1, B=1, C=−1, D=1 5)

4x2−2x+1		1		1		−x+1
	=		+		+
(x−1)2*(x2+x+1)		x−1		(x−1)2		x2+x+1)

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