(2x^2-10x+4)/(x^3-3x^2-x+3)=(2x²-10x+4)/(x-3)(x+1)(x-1)
设(2x²-10x+4)/(x-3)(x+1)(x-1)=A/(x-3)+B/(x+1)+C/(x-1)
则,2x²-10x+4=A(x²-1)+B(x²-4x+3)+C(x²-2x-3)
= (A+B+C)x²-(4B+2C)x+(-A+3B-3C)
A+B+C=2
4B+2C=10
-A+3B-3C=4
联立方程解得 A=-1 ,B=2 ,C=1
所以,(2x^2-10x+4)/(x^3-3x^2-x+3)化为分项分式为:
(2x^2-10x+4)/(x^3-3x^2-x+3)=-1/(x-3)+2/(x+1)+1/(x-1)