1、
2/(x-2)+kx/(x+2)(x-2)=3/(x+2)
两边乘(x+2)(x-2)
2(x+2)+kx=3(x-2)
增根即分母为0
(x+2)(x-2)=0
x=2,x=-2
x=2
代入2(x+2)+kx=3(x-2)
8+2k=0
k=-4
x=-2
代入2(x+2)+kx=3(x-2)
0-2k=-12
k=6
所以k=-4,k=6
2、
原式=[(a-2)/(a²+2)-(a-1)(a+2)²]×[(a+2)/(a-4)]
=[(a-2)/(a²+2)-(a-1)(a+2)²]×[(a+2)/(a-4)]
这里没写错吗?