问题标题：
先化简,再求职①x²-4x+4/x²-4-x+2/x-2其中x=根号2②（x-1/3x-x+1/x)×x/x²-1其中,x=2
问题描述：
先化简,再求职①x²-4x+4/x²-4-x+2/x-2其中x=根号2②（x-1/3x-x+1/x)×x/x²-1其中,x=2
陈继光回答：
(x²-4x+4)/(x²-4)-(x+2)/(x-2)
=(x-2)²/(x-2)(x+2)-(x+2)/(x-2)
=(x-2)²-(x+2)²/(x+2)(x-2)
=(x²-4x+4-x²-4x-4)/(x+2)(x-2)
=-8x/(x²-4)
=-8√2/［(√2)²-4］
=4√2
［(x-1)/3x-(x+1)/x)］*x/(x²-1)
=［(x-1)/3x-(x+1)/x)］*x/(x-1)(x+1)
=［(x-1-3x-3)/3x］*x/(x-1)(x+1)
=［(-2x-4)/3x］*x/(x-1)(x+1)
=(-2x-4)/(3x²-3)
=(-2*2-4)/(3*2²-3)
=-8/9
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