Question

一道二次根式化简题！！！十万火急!

根号a-根号b/根号a+根号b + （根号a/a+根号ab - 根号b/b-根号ab)÷1/根号b
有解题过程！！好的追加分！！

Answer 1

原式=(√a-√b)²/(√a-√b)(√a+√b)+[√a/√a(√a+√b)-√b/√b(√b-√a)]÷1/√b
=(a-2√ab+b)/(a-b)+[1/(√a+√b)+1/(√a-√b)]÷1/√b
=(a-2√ab+b)/(a-b)+{[(√a-√b)+(√a+√b)]/(√a+√b)(√a-√b)}×√b
=(a-2√ab+b)/(a-b)+[2√a/(a-b)]×√b
=(a-2√ab+b+2√ab)/(a-b)
=(a+b)/(a-b)