把3x^2+1/x^4-1化成分子中不含x的若干个分式的和小箭头是多少的平方的意思.
问题描述：
把3x^2+1/x^4-1化成分子中不含x的若干个分式的和
小箭头是多少的平方的意思.
答
（1/(x^2+1)）+（2/(x^2-1)）
答
3/x^(-2)+1/x^4-1
答
=3X^2+x^-4
答
(3x^2+1)/(x^4-1)=?
x^4-1=(x-1)(x+1)(xx+1)
设
(3x^2+1)/(x^4-1)=A/(x-1)+B/(x+1)+C/(xx+1)
(3x^2+1)=A(x+1)(xx+1)+B(x-1)(xx+1)+C(xx-1)
=A(xxx+xx+x+1)+B(xxx-xx+x-1)+C(xx-1)
=(A+B)xxx+(A-B+C)xx+(A+B)x+(A-B-C)
A+B=0
A-B+C=3
A-B-C=1
C=1
A=1
B=-1
(3x^2+1)/(x^4-1)=1/(x-1)-1/(x+1)+1/(xx+1)
答
LZ的题目是不是括号没括出来?(3x^2+1)/(x^4-1) 如果是这样,解题如下：原式=(3x^2+1)/(x^2+1)(x^2-1) =[3(x^2+1)-2]/(x^2+1)(x^2-1) =3/(x^2-1)-2/(x^2+1)(x^2-1) =(3/2)[1/{x-1)-1/(x+1)]-(1/2)[1/(x-1)-1/(x+1)]+1/...