Question

Rút gọn biểu thức : \(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}-\frac{b}{b-\sqrt{ab}}+\frac{a}{\sqrt{ab}+a}\right)-\frac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{2}\) với a,b > 0 \(a\ne b\)

Answer 1

\(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}-\frac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}\right)-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)

\(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}+\frac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{a-b}+\frac{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\right)-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)

\(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b+\sqrt{ab}+b+a-\sqrt{ab}}{a-b}\right)-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)

\(P=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{2\left(a+b\right)}{a-b}\right)-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)

\(P=\frac{\sqrt{a}-\sqrt{b}}{2}-\frac{\left|\sqrt{a}-\sqrt{b}\right|}{2}\)

TH1: \(a>b\Rightarrow P=\frac{\sqrt{a}-\sqrt{b}}{2}-\frac{\sqrt{a}-\sqrt{b}}{2}=0\)

TH2: \(0< a< b\Rightarrow P=\frac{\sqrt{a}-\sqrt{b}}{2}-\frac{\sqrt{b}-\sqrt{a}}{2}=\sqrt{a}-\sqrt{b}\)