先化简再求值 (x²-4/x²-4x+4)-(x-2/x+2),其中x=√2
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先化简再求值 (x²-4/x²-4x+4)-(x-2/x+2),其中x=√2
无语,请你把括号都标好:
如果没理解错,应该是这样的小孩子:
[(x-2)(x+2)]/[(x-2)^2]-(x-2)/(x+2)
=(x+2)/(x-2)-(x-2)/(x+2)
=[(x+2)^2-(x-2)^2]/(x^2-2^2)=4x/(x^2-4),带入x的只就可以了
(x^2-4)/(x^2-4x+4)-(x-2)/(x+2)=(x+2)(x-2)/(x-2)^2-(x-2)/(x+2)
=(x+2)/(x-2)-(x-2)(/(x+2)
=[(x+2)^2-(x-2)^2]/(x+2)(x-2)
=8x/[x^2-4]=8根号2/(2-4)=-4根号2
(x²-4/x²-4x+4)-(x-2/x+2)=(x+2/x-2)-(x-2/x+2)=8x/(x²-4)=8√2/(2-4)=-4 √2
(x²-4/x²-4x+4)-(x-2/x+2)
=(x+2)(x-2)/(x-2)²-(x-2/x+2)
=(x+2)/(x-2)-(x-2)/(x+2)
=(x+2)²/[(x-2)(x+2)]-(x-2)²/[(x-2)(x+2)]
=[(x+2)²-(x-2)²]/[(x-2)(x+2)]
=4x/(x²-4)
=4√2/(2-4)
=-2√2
=(x+2)/(x-2)-(x-2)/(x+2)
=8x/(x²-4)
=8√2 /(2-4)
=-4√2
楼上做错了。。倒数第三步应该是8x/(x²-4)=8√2/(2-4)= -4√2
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