①[1-1/(x-1)][x/(x+1)+1]÷[1-3x/(1-x^2)] ②计算:[x^2-4/(x^2-4x+4)

①[1-1/(x-1)][x/(x+1)+1]÷[1-3x/(1-x^2)] ②计算:[x^2-4/(x^2-4x+4)-x-2/(x+2)]÷x/(x-2)
③2/(a+1)-a-2/(a^2-1)÷a^2-2a/(a^2-2a+1)
totomm 1年前 已收到2个回答 举报

芝麻 花朵

共回答了20个问题采纳率:80% 举报

①[1-1/(x-1)][x/(x+1)+1]÷[1-3x/(1-x^2)]
=[(x-1-1)/(x-1)][(x+x+1)/(x+1)]÷[(1-x^2-3x)/(1-x^2)]
=[(x-2)/(x-1)][(2x+1)/(x+1)]÷[(1-x^2-3x)/(1-x^2)]
=[(2x^2+x-4x-2)/(x^2-1)]*[(1-x^2)/(1-x^2-3x)]
=[(2x^2-3x-2)/(x^2-1)]*[-(x^2-1)/(1-x^2-3x)]
=-(2x^2-3x-2)/(1-x^2-3x)
=(2x^2-3x-2)/(x^2+3x-1)
②计算:[x^2-4/(x^2-4x+4)-x-2/(x+2)]÷x/(x-2)
=[(x+2)(x-2)/(x-2)^2-x-2/(x+2)]÷x/(x-2)
=[(x+2)/(x-2)-x-2/(x+2)]÷x/(x-2)
=[((x+2)^2-(x-2)^2)/(x-2)(x+2)] *(x-2)/x
=[((x+2)+(x-2))((x+2)-(x-2))/(x-2)(x+2)] *(x-2)/x
=[2x*4/(x-2)(x+2)] *(x-2)/x
=8/(x+2)
③2/(a+1)-a-2/(a^2-1)÷a^2-2a/(a^2-2a+1)
=2/(a+1)-a-2/(a+1)(a-1)* (a-1)^2/a(a-2)
=2/(a+1)-(a-1)/a(a+1)
=(2a-(a-1))/a(a+1)
=(a+1)/a(a+1)
=1/a

1年前

4

uoldbixq 幼苗

共回答了31个问题 举报

1) [1-1/(x-1)][x/(x+1)+1]÷[1-3x/(1-x^2)]
=[(x-2)/(x-1)][(2x+1)/(x+1)]÷[(x^2+3x-1)/(x^2-1)]
=[(2x^2-3x-2)/(x^2-1)]÷[(x^2+3x-1)/(x^2-1)]
=(2x^2-3x-2)/(x^2+3x-1)

1年前

0
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