a.
\(M=\dfrac{a+1}{\sqrt{a}}+\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(a-1\right)\left(a+1\right)-\sqrt{a}\left(a-1\right)}{\sqrt{a}\left(a-1\right)}\)
\(=\dfrac{a+1}{\sqrt{a}}+\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{\left(a-1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)}\)
\(=\dfrac{a+1}{\sqrt{a}}+\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\)
\(=\dfrac{a+1+a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}=\dfrac{a+2\sqrt{a}+1}{\sqrt{a}}\)
b.
\(M=\dfrac{a-2\sqrt{a}+1+4\sqrt{a}}{\sqrt{a}}=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}+4\)
Do \(\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}\ge0\Rightarrow M\ge4\)
Dấu "=" xảy ra khi \(\sqrt{a}-1=0\Rightarrow a=1\) ko thỏa mãn ĐKXĐ
\(\Rightarrow M>4\)
