已知a²+2a-√3=0,求代数式[（a-2)/（a²+2a)-(a-1)/（a²+4a+

由[（a-2)/（a（a+2)-（a-1)/（a+2)²]÷（a-4)/（a+2)
=[（a-2)(a+2)/a(a+2)²-a(a-1)/a(a+2)²
=(a-4)/a(a+2)²×(a+2)/(a-4)
=1/a(a+2)
∵a²+2a=√3,
∴1/a(a+2)
=1/(a²+2a)
=1/√3
=√3/3.