a. ĐKXĐ \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
b.
\(P=\left(\frac{1}{\sqrt{a}-1}+\frac{1}{\sqrt{a}+1}\right):\frac{\sqrt{a}+1}{a-1}\\ =\left(\frac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a}+1}{a-1}\\ =\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+1}=\frac{2\sqrt{a}}{\sqrt{a}+1}\)
c. \(\sqrt{a}=\sqrt{6+2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
\(P=\frac{2\sqrt{a}}{\sqrt{a}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1+1}=\frac{2\sqrt{5}+2}{\sqrt{5}+2}\)(nếu rút gọn đc thì bạn tiếp tục nha)
d.
\(P=\frac{2\sqrt{a}}{\sqrt{a}+1}=\frac{2}{5}\Leftrightarrow2\sqrt{a}=\frac{2}{5}\left(\sqrt{a}+1\right)\\ \Leftrightarrow5\sqrt{a}=\sqrt{a}+1\Leftrightarrow4\sqrt{a}=1\Leftrightarrow\sqrt{a}=\frac{1}{4}\\ \Rightarrow a=\frac{1}{16}\)
