`a)`\(M=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(M=\left[\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{1}{\sqrt{a}-1}\right]:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
`b)`\(M=-1\)
\(\Leftrightarrow\dfrac{\sqrt{a}-1}{\sqrt{a}}=-1\)
\(\Leftrightarrow\sqrt{a}-1=-\sqrt{a}\)
\(\Leftrightarrow a=\dfrac{1}{4}\) ( tm )
`c)`\(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(M=1-\dfrac{1}{\sqrt{a}}\)
Ta có: \(\sqrt{a}>0\)
\(\Rightarrow\dfrac{1}{\sqrt{a}}>0\)
\(\Rightarrow M< 1-0=1\)
`d)`\(M< 0\)
\(\Leftrightarrow1-\dfrac{1}{\sqrt{a}}< 0\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}}>1\)
\(\Leftrightarrow1>\sqrt{a}\)
\(\Leftrightarrow a< 1\)
Vậy \(S=\left\{a|0< a< 1\right\}\)
