先化简,在求值:(3x/x+1-x/x-1)÷x-2/x²-1,其中选√3/2.

显示全部楼层 · 2012-7-25 21:05:54

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千问 · 2012-7-25 21:05:54

把x-2/x2-1分解因式得x-2/（x+1)(x-1)再把分解过的因式的倒数乘进去得（3x/x+1）*（x+1)(x-1)/（x-2）-（x/x-1）*（x+1)(x-1)/（x-2）得=3x(x-1)/(x-2)-x(x+1)/(x-2) =[3x(x-1)-x(x+1)]/(x-2) =2x把x的值代进去即可...

千问 · 2012-7-25 21:05:54

(3x/x+1-x/x-1)÷x-2/x2-1=(3x/x+1-x/x-1)*(x-1)(x+1)/(x-2)=x[3(x-1)-(x+1)]/(x-2)=x(3x-3-x-1)/(x-2)=x(2x-4)/(x-2)=2x(x-2)/(x-2)=2x=2*√3/2=√3...

千问 · 2012-7-25 21:05:54

=[3x(x-1)-x(x+1)]/[(x+1)(x-1)]/[(x-2)/(x^2-1)]=2x(x-2)/(x-1)^2*(x^2-1)/(x-2)=2x=2*根号3/2=根号3...