\(1.ĐKXĐ\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
\(M=\frac{a+1}{\sqrt{a}}+\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(\Leftrightarrow M=\frac{a+1}{\sqrt{a}}+\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\)
\(\Leftrightarrow M=\frac{a+2\sqrt{a}+1}{\sqrt{a}}\)
2.
có \(M=\frac{a+2\sqrt{a}+1}{\sqrt{a}}\)
\(\Leftrightarrow M=4+\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}\)
vì \(a>0\) và \(a\ne0\)
\(\Rightarrow\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>0\Leftrightarrow4+\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>4\)
Vậy M > 4
