当x=根号3/3,求代数式(x^2-3x)/(x-2)-[x+3/(2-x)]的值
问题描述：
当x=根号3/3,求代数式(x^2-3x)/(x-2)-[x+3/(2-x)]的值
答
-(17+sqrt3)/11
答
原代数式=x(x-3)/(x-2)-[x-3/(x-2)]=-x+(x²-3x-3)/(x-2)=(-x²+2x+x²-3x+3)/(x-2)=(3-x)/(x-2)
将x=√3/3代入
原式=(3-√3/3)/(√3/3-6)=-(17+√3)/11
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