Question

已知A/x+1+B/x-3=x-3/(x+1)(x-1),求A,B的值

Answer 1

A/x+1+B/x-3=[A(x-3)+B(x+1)]/[(x+1)(x-3)]=[(A+B)x+B-3A]/[(x+1)(x-3)]=x-3/(x+1)(x-1),
则)[(A+B)x+B-3A]*(x-1)=(x-3)*(x-3),
即(A+B)x²-(A+B)x+(B-3A)x-B+3A=x²-6x+9
即(A+B)x²-4Ax-B+3A=x²-6x+9
所以A+B=1，4A=6,-B+3A=9
无解。
不知道你是不是分母写错了，再问吧。

Answer 2

A/ (x+1) +B/(x-3)=(x-3)/(x+1)(x-1),
[A(x-3)+B(x+1)]/(x-3)(x+1)=(x-3)/(x+1)(x-1)
[A(x-3)+B(x+1)]/(x-3)=(x-3)/(x-1)
A(x-3)(x-1)+B(x+1)(x-1) = (x-3)²
A(x²-4x+3)+B(x²-1)= x²-6x+9
Ax²-4Ax+3A+Bx²-B= x²-6x+9
x²(A+B)-4Ax+3A-B= x²-6x+9
由上式可知
A+B=1
4A=6
3A-B=9
可得：
A=3/2,B=-1/2和B=-9/2
A=5/2,A=3/2和B=-3/2
则方程不平衡题目有错
若A/ (x+1) +B/(x-1)=(x-3)/(x+1)(x-1),
[A(x-1)+B(x+1)]/(x-1)(x+1)=(x-3)/(x+1)(x-1)
A(x-1)+B(x+1)=(x-3)
A(x-3)(x-1)+B(x+1)(x-1) = (x-3)²
x(A+B)+B-A= x-3
由上式可知
A+B=1
B-A=-3
A=2，B=-1