(x²+x-3)/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3),
4个回答

(x²+x-3)/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3),

求A,B,C的值.

分别通分得

A/(x-1)=A(x-2)(x-3)/ (x-1)(x-2)(x-3)-------(1)

B/(x-2)=B(x-1)(x-3)/ (x-1)(x-2)(x-3)-------(2)

C/(x-3)=C(x-1)(x-2)/ (x-1)(x-2)(x-3)-------(3)

则上面三式有,

x²+x-3= A(x-2)(x-3)+ B(x-1)(x-3)+ C(x-1)(x-2)

x²+x-3=(A+B+C)x²-(5A+4B+3C) x+(6A+3B+2C)

所以,

A+B+C=1--------(4)

5A+4B+3C=-1----(5)

6A+3B+2C=-3----(6)

联立(4)(5)(6)得 A,B,C的值.,计算得A=-1/2,B=-3,C=9/2.