已知2x+3/x(x-1)(x+2)=A/x+B/x-1+C/x+2,A,B,C为常数,求A,B,C

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(2x+3)/[x(x-1)(x+2)]=A/x+B/(x-1)+C/(x+2)=[A(x²+x-2)+B(x²+2x)+C(x²-x)]/[x(x-1)(x+2)]=[(A+B+C)x²+(A+2B-C)x-2A]/[x(x-1)(x+2)]于是A+B+C=0A+2B-C=2-2A=3联立解得A=-3/2B=5/3C=-1/6