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(x/x^3-1-1/x^2+x+1)除以(1/x-1-x/x^2+x+1)其中x=1/2
题目详情
(x/x^3-1-1/x^2+x+1)除以(1/x-1-x/x^2+x+1) 其中x=1/2
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答案和解析
(x/x^3-1-1/x^2+x+1)除以(1/x-1-x/x^2+x+1)
=[x/(x-1)(x^2+x+1)-1/(x^2+x+1)]÷[1/(x-1)-x/(x^2+x+1)]
=[x/(x-1)(x^2+x+1)-(x-1)/(x^2+x+1)]÷[(x^2+x+1)/(x-1)(x^2+x+1)-(x-1)x/(x-1)(x^2+x+1)]
=[x-(x-1)]/(x-1)(x^2+x+1)÷[(x^2+x+1)-(x-1)x]/(x-1)(x^2+x+1)
=1/(x-1)(x^2+x+1)÷（2x+1）/(x-1)(x^2+x+1)
=1/(2x+1)
当x=1/2时,原式=1/(2x+1)=1/2
=[x/(x-1)(x^2+x+1)-1/(x^2+x+1)]÷[1/(x-1)-x/(x^2+x+1)]
=[x/(x-1)(x^2+x+1)-(x-1)/(x^2+x+1)]÷[(x^2+x+1)/(x-1)(x^2+x+1)-(x-1)x/(x-1)(x^2+x+1)]
=[x-(x-1)]/(x-1)(x^2+x+1)÷[(x^2+x+1)-(x-1)x]/(x-1)(x^2+x+1)
=1/(x-1)(x^2+x+1)÷（2x+1）/(x-1)(x^2+x+1)
=1/(2x+1)
当x=1/2时,原式=1/(2x+1)=1/2
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