Question

化简：〔1/a的平方＋2a-1/a的平方＋4a+4〕÷2/a的平方＋2a,并求出a=根号3时代数式的值

Answer 1

[1/(a²+2a) -1/(a²+4a+4)]÷[2/(a²+2a)]

=[1/a(a+2) -1/(a+2)²]÷[2/a(a+2)]

=[1/a(a+2) -1/(a+2)²]×[a(a+2)/2]

=(1/2) -a/[2(a+2)]

=(1/2) -(a+2-2)/[2(a+2)]

=(1/2) -(1/2) +1/(a+2)

=1/(a+2)

=1/(2+√3)

=(2-√3)/[(2+√3)(2-√3)]

=2-√3