ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
a) \(M=\) \(\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(M=\frac{\sqrt{a}-1}{\sqrt{a}}\)
b) Xét hiệu: \(M-1\)
\(M-1=\frac{\sqrt{a}-1}{\sqrt{a}}-1=\frac{\sqrt{a}-1}{\sqrt{a}}-\frac{\sqrt{a}}{\sqrt{a}}=\frac{-1}{\sqrt{a}}\)
Vì \(\left\{{}\begin{matrix}-1< 0\\\sqrt{a}\ge0\end{matrix}\right.\)
\(\Rightarrow\frac{-1}{\sqrt{a}}< 0\)\(\Leftrightarrow M-1< 0\Leftrightarrow M< 1\)
Vậy: \(M< 1\)
