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.
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![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.](/uploads/image/z/15239829-21-9.jpg?t=%E6%8A%8A3x%5E2%2B1%2Fx%5E4-1%E5%8C%96%E6%88%90%E5%88%86%E5%AD%90%E4%B8%AD%E4%B8%8D%E5%90%ABx%E7%9A%84%E8%8B%A5%E5%B9%B2%E4%B8%AA%E5%88%86%E5%BC%8F%E7%9A%84%E5%92%8C%E5%B0%8F%E7%AE%AD%E5%A4%B4%E6%98%AF%E5%A4%9A%E5%B0%91%E7%9A%84%E5%B9%B3%E6%96%B9%E7%9A%84%E6%84%8F%E6%80%9D.)
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.
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.
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.
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/(x^2+1)
=1/(x-1)-1/(x+1)+1/(x^2+1)
这个是不是你要的结果?
(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+...
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(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)
收起
(1/(x^2+1))+(2/(x^2-1))
3/x^(-2)+1/x^4-1
=3X^2+x^-4