a) \(A=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left[\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}\)
b) Ta có : \(\frac{1}{\sqrt{a}}>0\Leftrightarrow-\frac{1}{\sqrt{a}}< 0\Rightarrow\) \(A=1-\frac{1}{\sqrt{a}}< 1\)
a) \(A=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}=\frac{\sqrt{a}-1}{\sqrt{a}}\)
b) Do \(\sqrt{a}\ge0\) => \(\sqrt{a}-1< \sqrt{a}\)=> \(\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)
