√(a-5)+b^2-8b+16 =0√(a-5)+(b–4)^2 =0 （1）因为 √(a-5)≧0,(b–4)^2≧0 ,要使（1）式有解,必须 √(a-5)=0 且 (b–4)^2=0解得：a=5,b=4而√ab/(√a+√b)= (√ab*(√a-√b) )/(a-b)=(√(5*4)*(√5-√4) )/(5-4)=...谢谢喽