Bài 1:
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)
\(M=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
b: Thay \(a=22-12\sqrt{2}\) vào M, ta được:
\(M=\dfrac{3\sqrt{2}-2-1}{3\sqrt{2}-2}=\dfrac{12-3\sqrt{2}}{14}\)