\(a,P=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right).\frac{1}{2a\sqrt{a}}\left(a>0;a\ne1\right)\)
\(=\left[\frac{\left(\sqrt{a}+1\right)^2}{a-1}-\frac{\left(\sqrt{a}-1\right)^2}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right].\frac{1}{2a\sqrt{a}}\)
\(=\frac{a^2+2\sqrt{a}+1-\left(a^2-2\sqrt{a}+1\right)+4a\sqrt{a}-4\sqrt{a}}{a-1}.\frac{1}{2a\sqrt{a}}\)
\(=\frac{4a\sqrt{a}}{\left(a-1\right).2a\sqrt{a}}=\frac{2}{a-1}\left(đpcm\right)\)
\(b,P=a\Leftrightarrow\frac{2}{a-1}=a\Leftrightarrow a^2-a=2\)
\(\Leftrightarrow a^2-a-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-1\left(ktm\right)\end{matrix}\right.\)
Vậy \(a=2\)
