待定系数法拆项

Question

x/((x-3)(x+1))=1/4*[3/(x-3)+1/(x+1)]能不能具体解答一下用
a[(b/(x-3)+c/(x+1))]怎么得到上面这个结果，演示一下过程。谢谢

Answer 1

a[(b/(x-3)+c/(x+1))]
=a[(bx+b)+(cx-3c)]/((x-3)(x+1))
=x/((x-3)(x+1))

a(b+c)=1
a(b-3c)=0

解得，b=3c
4ac=1
a=1/4, c=1, b=3

所以：
x/((x-3)(x+1))=1/4*[3/(x-3)+1/(x+1)]

Answer 2

b/(x-3)+c/(x+1))] = (b(x+1) + c(x-3))/(x+1)(x-3)
=((b+c)x + b - 3c)/(x+1)(x-3)

于是 b+c = 1
b-3c = 0

b = 3/4, c = 1/4
a没有用的

Answer 3

x/((x-3)(x+1))=1/4[(x+1)-(x-3)]/(x-3)(x+1)==1/4[(x+1)/(x-3)(x+1)-(x-3)
/(x-3)(x+1)]=1/4[1/(x-3)-1/x+1)]