已知60/(x+1)(x-2)(x+3) = A/x+1 + B/x-2 + C/x+3 且A,B,C为常数,求A+B+C=?

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已知60/(x+1)(x-2)(x+3)=A/x+1+B/x-2+C/x+3且A,B,C为常数,求A+B+C=?已知60/(x+1)(x-2)(x+3)=A/x+1+B/x-2+C/x+3且A,B,C

已知60/(x+1)(x-2)(x+3) = A/x+1 + B/x-2 + C/x+3 且A,B,C为常数,求A+B+C=?
已知60/(x+1)(x-2)(x+3) = A/x+1 + B/x-2 + C/x+3 且A,B,C为常数,求A+B+C=?

已知60/(x+1)(x-2)(x+3) = A/x+1 + B/x-2 + C/x+3 且A,B,C为常数,求A+B+C=?
A/x+1 + B/x-2 + C/x+3
=[A(x-2)(x+3)+B(x+1)(x+3)+C(x+1)(x-2)]/[(x+1)(x-2)(x+3)]
=60/(x+1)(x-2)(x+3)
因此A(x-2)(x+3)+B(x+1)(x+3)+C(x+1)(x-2)=60（恒等于）
A(x^2+x-6)+B(x^2+4x+3)+C(x^2-x-2)=60
(A+B+C)x^2+(A+4B-C)x+(-6A+3B-2C)=60
因为上式恒等,所以二次项系数为0.因此
A+B+C=0

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