Question

计算 [√a-√b/√b+a-b/√a*√a+√b/a+2√ab+b]/(√a+√b)

Answer 1

[√a-√b/√b+a-b/√a*√a+√b/a+2√ab+b]/(√a+√b)因为√b/√b=1，b/√a*√a=b。所以
= [√a-1+a-b+√b/a+2√ab+b]/(√a+√b)
= [√a-1+a+√b/a+2√ab]/(√a+√b)
= [a+√a+2√ab+√b/a-1]/(√a+√b)
= [√a(√a+1+2b)+√b/a-1]/(√a+√b)
= [√a(√a+2b+1)+√b/a-1]/(√a+√b)

Answer 2

原式=[(√a-√b)/√b+(√a+√b)(√a-√b)/√a×(√a+√b)/(√a+√b)²]/(√a+√b)
=[(√a-√b)/√b+(√a+√b)(√a-√b)/√a×1/(√a+√b)]/(√a+√b)
=[(√a-√b)/√b+(√a-√b)/√a]/(√a+√b)
=(√a/√b-1+1-√b/√a)/(√a+√b)
=[(a-b)/(√a*√b)]/(√a+√b)
=[(√a+√b)(√a-√b)/(√a*√b)]/(√a+√b)
=(√a-√b)/(√a*√b)