(a平方+a/a平方+2a+1)÷(1+1/a-1),其中a=tan60°-根号2sin45°

题目

(a平方+a/a平方+2a+1)÷(1+1/a-1),其中a=tan60°-根号2sin45°

答案

a=tan60°-根号2sin45°
=√3-√2*1/2√2
=√3-1
(a平方+a/a平方+2a+1)÷(1+1/a-1),
=a(a+1)/(a+1)平方÷[a/(a-1)]
=a/(a+1)*(a-1)/a
=(a-1)/(a+1)
=(√3-1-1)/(√3-1+1)
=(√3-2)/√3
=1-2/3√3