Question

化简： [√a+(b-√ab)/(√a+√b)]÷[a/(√ab+b)+b/(√ab-a)-(a+b)/√ab]

Answer 1

[√a+(b-√ab)/(√a+√b)]÷[a/(√ab+b)+b/(√ab-a)-(a+b)/√ab]
=[√a+(b-√ab)/(√a+√b)]÷[a/(√ab+b)+b/(√ab-a)-(a+b)/√ab]
=[a+√ab/(√a+√b)+(b-√ab)/(√a+√b)]÷[a/(√ab+b)+b/(√ab-a)-(a+b)/√ab]
=[a+b/(√a+√b)]÷[(√ab-1)(a+b)/√ab]
=√ab/(√ab-1)(√a+√b)

追问

√a+(b-√ab)/(√a+√b)→[a+√ab/(√a+√b)+(b-√ab)/(√a+√b) 这一步转化我看不懂 能详细解释一下么

追答

好想错了...
分子：
√a+(b-√ab)/(√a+√b)=(a+√ab+b-√ab)/(√a+√b)=(a+b)/(√a+√b)=(a+b)(√b-√a)/(b-a)

分母：
a/(√ab+b) + b/(√ab-a) - (a+b)/√ab...........第一分式同除a、第二分式同除b--->
=1/(√b/a+b/a) + 1/(√a/b-a/b) -(a+b)/√ab..............合并第一、第二分式--->
=(√b/a+b/a + √a/b-a/b) / [(√b/a+b/a) *(√a/b-a/b)] -(a+b)/√ab
=(√b/a+b/a + √a/b-a/b) /(√b/a- √a/b) -(a+b)/√ab...................第一个分式分子分母同乘√ab--->
=(b+b√b/√a+a-a√a/√b)/(b-a) -(a+b)/√ab
=(a+b)/(b-a)+(b²-a²)/√ab/(b-a) -(a+b)/√ab
=(a+b)/(b-a)+(b+a)/√ab -(a+b)/√ab
=(a+b)/(b-a)

原式：
=分子/分母-√b
=[(a+b)(√b-√a)/(b-a)] / [(a+b)/(b-a)] -√b
=√b-√a -√b
=-√a