（1）
﹛x-2-[12/(x+2)]﹜÷(4-x)/(x+2)
=[x-2-12/(x+2)]÷[(4-x)/(x+2)]
={ [(x-2)(x+2)-12]/(x+2) }×[(x+2)/(4-x)]
=[(x²-4-12)/(x+2)]×[-(x+2)/(x-4)]
=(x²-16)×[-1/(x-4)]
=(x+4)(x-4)×[-1/(x-4)]
=-(x+4)
=-x-4
（2）
3/(x-1)-(x+3)/(x^2-1)=0
3(x+1)/(x-1)(x+1)-(x+3)/(x-1)(x+1)=0
[3(x+1)-(x+3)]/(x-1)(x+1)=0
(3x+3-x-3)/(x-1)(x+1)=0
2x/(x-1)(x+1)=0
2x=0
x=0
經(jīng)檢驗(yàn)x=0是方程的解