右边通分=[A(x-2)(x-3)+B(x-3)+C(x-2)]/(x-2)(x-3)=[Ax²+(B+C-5A)x-(6A+3B+2C)]/(x-2)(x-3)=(x²-1)/(x-2)(x-3)对应项系数相等,得：A=1B+C-5A=06A+3B+2C=1解方程组,得：A=1,B=-15,C=20不懂追问~ 希望我的回...