a) \(ĐKXĐ:\hept{\begin{cases}a>0\\b>0\\a\ne b\end{cases}}\)
\(A=\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)\)
\(\Leftrightarrow A=\frac{a+\sqrt{ab}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\frac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}-\frac{b}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{a+b}{\sqrt{ab}}\right)\)
\(\Leftrightarrow A=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\frac{a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(a-b\right)}{\sqrt{ab}\left(a-b\right)}\)
\(\Leftrightarrow A=\left(\sqrt{a}-\sqrt{b}\right)\cdot\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{a^2-a\sqrt{ab}-b\sqrt{ab}-b^2-a^2+b^2}\)
\(\Leftrightarrow A=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{-a\sqrt{ab}-b\sqrt{ab}}\)
\(\Leftrightarrow A=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{-\sqrt{ab}\left(a+b\right)}\)
\(\Leftrightarrow A=\frac{-\sqrt{a}-\sqrt{b}}{a+b}\)
b) Thay \(a=6-2\sqrt{5}\)và \(b=5\)vào A ta được :
\(A=\frac{-\sqrt{6-2\sqrt{5}}-\sqrt{5}}{6-2\sqrt{5}+5}=\frac{-\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}}{1-2\sqrt{5}}=\frac{1-2\sqrt{5}}{1-2\sqrt{5}}=1\)
Vậy ...
