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分式的加减计算:(1/x-1)+(1/x+1)-(2x/x^2+1)-(4x/x^4+1)-(8x/x^8+1) 注:“/”指除号.
人气:450 ℃ 时间:2019-10-19 13:34:32
解答
(1/x-1)+(1/x+1)-(2x/x^2+1)-(4x/x^4+1)-(8x/x^8+1)
=(x+1+x-1)/(x^2-1)-(2x/x^2+1)-(4x/x^4+1)-(8x/x^8+1)
=[2x(x^2+1)-2x(x^2-1)]/(x^4-1)-(4x/x^4+1)-(8x/x^8+1)
=[4x(x^4+1)-4x(x^4-1)]/(x^8-1) )-(8x/x^8+1)
=[8x(x^8+1)-8x(x^8-1)]/(x^16-1)
=16x/(x^16-1)
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