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还是解代数题
等式的 分子:[(x-2)/(x+2)加上(x-1)/(x+1)]
分母:{[x/(x+1)]减去[(2x-3)/x]}
还是要有过程啊..
嗷嗷嗷
人气:473 ℃ 时间:2020-05-23 18:08:34
解答
(x-2)/(x+2)+(x-1)/(x+1)
=[(x-2)(x+1)+(x-1)(x+2)]/(x+1)(x+2)
=2x^2/(x+1)(x+2)
x/(x+1)-(2x-3)/x
=(x^2-2x^2+3x-2x+3)/[x(x+1)]
=(3+x-x^2)/x(x+1)
原式
=2x^3/(x+2)(3+x-x^2)
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