已知：X的平方-1/（X-2）（X-3)=A+B/(X-2)+C/X-3,求ABC的值

(x^2-1)/{(x-2)(x-3) } = A + B/(x-2) + C(x-3)
等式右边统分：
{A(x-2)(x-3) + B(x-3) + C(x-2)} / {(x-2)(x-3)}
= {A(x^2-5x+6) + B(x-3) + C(x-2)} / {(x-2)(x-3)}
= {Ax^2-5Ax+6A + Bx-3B + Cx-2C} / {(x-2)(x-3)}
= {Ax^2+(-5A+B+C)x+(6A-3B-2C)} / {(x-2)(x-3)}
分母与等式左边一致,将分子与等式左边分子比较得：
A=1
-5A+B+C=0
6A-3B-2C=-1
解得：A=1,B=-3,C=8