初二分式的化简

题目

初二分式的化简
①若x=3,求 (1+x)/(x²+x-2)÷（ x-2+(3)/(x+2) )的值
②使分式 （ x绝对值—1）/（x-3)(x+1)有意义,x取何值
③已知abc=1,a+b+c=2,a²+b²+c²=3 ；求 1/（ab+c-1) + 1/(bc+a-1) + 1/(ca+b-a) 的值
④已知 M/（x²-y²）=（2xy-y²）/(x²-y²）+（x-y)/(x+y),则M=?
③已知abc=1，a+b+c=2,a²+b²+c²=3 ；求 1/（ab+c-1) + 1/(bc+a-1) + 1/(ca+b-1 )的值
最后一项应该是-1

答案

(1)
(1+x)/(x²+x-2)÷[（x-2)+3/(x+2)]
=[(x+1)/(x^2+x-2)]/[(x^2-1)/(x+2)]
=(x+1)(x+2)/[(x^2+x-2)(x^2-1)]
=(x+1)(x+2)/[(x+2)(x+1)(x-1)^2]
=1/(x-1)^2
=1/(3-1)^2
=1/4
(2)
(|x|-1)/[(x-3)(x+1)],分式有意义,分母不等于0
(x-3)(x+1)≠0
x≠3且x≠-1
(3)
题目好像有点问题,最后一项分母是-a吗?
(4)
M/(x^2-y^2)=(2xy-y^2)/(x^2-y^2)+(x-y)/(x+y)
M=(2xy-y^2)+(x-y)^2
=2xy-y^2+x^2-2xy+y^2
=x^2