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分式计算(1/2-x)-[(x+2/x^2-2x)-(x-1/x^2-4x+4)]/(x-4/2x)
人气:357 ℃ 时间:2019-11-18 10:11:20
解答
(1/2-x)-[(x+2/x^2-2x)-(x-1/x^2-4x+4)]/(x-4/2x)
=-1/(x-2)-[(x+2)/x(x-2) - (x-1)/(x-2)^2]*2x/(x-4)
=-1/(x-2)-[((x+2)(x-2)-x(x-1))/x(x-2)^2]*2x/(x-4)
=-1/(x-2)-[(x^2-4-x^2+x)/x(x-2)^2]*2x/(x-4)
=-1/(x-2)-[(x-4)/x(x-2)^2]*2x/(x-4)
=-1/(x-2)- 2/(x-2)^2
=(-(x-2)-2)/(x-2)^2
=(-x+2-2)/(x-2)^2
=-x/(x-2)^2呃第二步通分为什么都要乘以(x-2)啊..[(x+2)/x(x-2) - (x-1)/(x-2)^2]公分母为x(x-2)^2
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