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当x=根号3/3,求代数式(x^2-3x)/(x-2)-[x+3/(2-x)]的值
人气:289 ℃ 时间:2019-09-22 09:43:38
解答
(x^2-3x)/(x-2)-[x+3/(2-x)]=(x^2-3x)/(x-2)+[-x+3/(x-2)]=(x^2-3x)/(x-2)+[(-x^2+2x+3)/(x-2)]=(x^2-3x-x^2+2x+3)/(x-2)=(-x+3)/(x-2)将x=√3/3代入原式=(-√3/3+3)/(√3/3-2)=-(17+√3)/11
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