已知a+4b=0,则代数式(1-（b/a+3b)）÷a²-2ab+b²/a²-b²的值为

a+4b=0
a=-4b
[1-b/(a+3b)]÷(a²-2ab+b²)/(a²-b²)
=[(a+3b)/(a+3b)-b/(a+3b)]÷(a-b)²/[(a-b)(a+b)]
=(a+3b-b)/(a+3b)÷(a-b)/(a+b)
=(a+2b)/(a+3b)*(a+b)/(a-b)
=(-4b+2b)/(-4b+3b)*(-4b+b)/(-4b-b)
=(-2b)/(-b)*(-3b)/(-5b)
=2*3/5
=6/5