将等式的右边通分得，Ax(x−1)+B(x−1)+Cx2x2(x−1)=(A+C)x2+(B−A)x−Bx2(x−1)，∵2x2+x−11x2(x−1)=Ax+Bx2+Cx−1，∴2x2+x−11x2(x−1)=(A+C)x2+(B−A)x−Bx2(x−1)，∴A+C＝2B−A＝1−B＝−11，解得A＝10B＝11C...