高考问答扫码软件辨别真伪:解一道数学题
来源：百度文库 编辑：高考问答 时间：2024/04/29 14:46:34
[1/X-(X-3)/(1-X)+2/(X^2-X)]除以(1+3/X-4/X^2)
除以(1+3/X-4/X^2) ，就是乘以它的倒数
(1+3/X-4/X^2)＝（X^2+3X-4）/X^2=(X-1)(X+4)/X^2
所以倒数为X^2/[(X-1)(X+4)]
原式＝1/X-(X-3)/(1-X)+2/(X^2-X)=1/X-(X-3)/(1-X)+2/[X(X-1)]
最后一个分式2/[X(X-1)]=-2*(X-1-X)/[X(X-1)]=-2*[1/X-1/(X-1)]
带入原式＝1/X-(X-3)/(1-X)-2/X+2/(X-1)=1/X-2/X+(X-3)/(X-1)+2(X-1)
=-1/X+(X-3+2)/(X-1)=-1/X+1=1-1/X=(X-1)/X
在乘以那个倒数 [(X-1)/X]* X^2/[(X-1)(X+4)]=X/(X+4)
[1/x+(x-3)/(x-1)+2/x(x-1)]/[(x^2+3x-4)/x^2]
=[(x-1)^2/x(x-1)]/[(x+4)(x-1)/x^2]
=[(x-1)/x]*[x^2/(x+4)(x-1)]
=x/(x+4)