a) \(M=\dfrac{\sqrt{a}}{2}\cdot\left(-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(a>0;a\ne1\right)\)
\(=\left(\dfrac{\sqrt{a}}{2}\cdot\dfrac{-1}{2\sqrt{a}}\right)\cdot\left[\dfrac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)
\(=\dfrac{-1}{4}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{-1}{4}\cdot\dfrac{\sqrt{a}\left[\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right]}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{-1}{4}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1-\sqrt{a}-1\right)\left(\sqrt{a}-1+\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{-1}{4}\cdot\dfrac{-2\sqrt{a}\cdot2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{-4}\cdot\dfrac{-4a}{a-1}\)
\(=\dfrac{a}{a-1}\)
b) Sửa: \(M\ge-2\)
\(\Leftrightarrow\dfrac{a}{a-1}\ge-2\)
\(\Leftrightarrow\dfrac{a}{a-1}+2\ge0\)
\(\Leftrightarrow\dfrac{3a-2}{a-1}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a< \dfrac{2}{3}\end{matrix}\right.\)
Theo đkxđ thì \(\Leftrightarrow\left[{}\begin{matrix}a>1\\0< a< \dfrac{2}{3}\end{matrix}\right.\)
