Question

待定系数法求解一题（x+1)(x+2)/-x+1=(X+1/A)+(x+2/B)

Answer 1

右边通分

=(x+1)(x+2)/[A(x+2)+B(x+1)]

=(x+1)(x+2)/[(A+B)x+(2A+B)]

=(x+1)(x+2)/(-x+1)

所以(A+B)x+(2A+B)=-x+1

所以A+B=-1

2A+B=1

相减

A=2,

B=-1-A=-3