Bài 2:
a) Ta có: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Ta có: \(P-\dfrac{1}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{3}\)
\(=\dfrac{\sqrt{a}-2-\sqrt{a}}{3\sqrt{a}}=\dfrac{-2}{3\sqrt{a}}< 0\forall a\) thỏa mãn ĐKXĐ
\(\Leftrightarrow P< \dfrac{1}{3}\)