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

问题描述:

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),
通分,得:
(x^2-1)/(x-2)(x-3)=[A(x-2)(x-3)+B(x-3)+C(x-2)]/(x-2)(x-3),
分母相同,则分子相等,所以
x^2-1=A(x^2-5x+6)+B(x-3)+C(x-2)=Ax^2+(-5A+B+C)*x+(6A-3B-2C),
多项式对应项的系数相等,所以
A=1,
-5A+B+C=0,
6A-3B-2C=-1,
解得:
A=1,
B=-3,
C=8。

(x^2-1)/(x-2)(x-3)
=A+B/(x-2)+C/(x-3)
=[A(x-2)(x-3)+B(x-3)+C(x-2)]/(x-2)(x-3)
=[A(x^2-5x+6)+Bx-3B+Cx-2C]/(x-2)(x-3)
=[Ax^2+(B+C-5A)x+6A-3B-2C]/(x-2)(x-3)
对比,得
A=1
B+C-5A=0
6A-3B-2C=-1
解得
A=1
B=-3
C=8

(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)}= ...