已知√x=√a+1/√a(0＜a＜1) ,求代数式(x^2+x-6)/x÷［(x+3)/(x^2-2x) ］-［x-2+

(x^2+x-6)/x÷〔(x+3)/(x^2-2x) 〕-〔x-2+√(x^2-4x) 〕/〔x-2-√(x^2-4x) 〕
=(x+3)(x-2)/x÷〔(x+3)/x(x-2) 〕-〔x-2+√(x^2-4x) 〕^2/(〔x-2-√(x^2-4x) 〕*(x-2+√(x^2-4x) 〕)
=(x-2)^2-〔x-2+√(x^2-4x) 〕^2/4
√x=√a+1/√a
x=a+1/a+2
代数式=（a+1/a)^2-(a+1/a+√(a+1/a+2)(a+1/a-2))^2/4
=（a+1/a)^2-a^2
=2+1/a^2