
当x=3时,求(x+2/x²-2x-x-1/x²-4x+4)÷x²-16/x²+4x的值 5
2个回答
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[(x+2)/(x²-2x)-(x-1)/(x²-4x+4)]÷[(x²-16)/(x²+4x)]
={(x+2)/[x(x-2)]-(x-1)/(x-2)²}×x(x+4)/[(x+4)(x-4)]
={(x-2)(x+2)/[x(x-2)²]-x(x-1)/[x(x-2)²]}×[x/(x-4)]
=[(x²-4-x²+x]/[x(x-2)²]×[x/(x-4)]
=(x-4)/[(x-2)²(x-4)]
=1/(x-2)²
=1/(3-2)²
=1
={(x+2)/[x(x-2)]-(x-1)/(x-2)²}×x(x+4)/[(x+4)(x-4)]
={(x-2)(x+2)/[x(x-2)²]-x(x-1)/[x(x-2)²]}×[x/(x-4)]
=[(x²-4-x²+x]/[x(x-2)²]×[x/(x-4)]
=(x-4)/[(x-2)²(x-4)]
=1/(x-2)²
=1/(3-2)²
=1
来自:求助得到的回答
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