Question

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

Answer 1

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