Đặt bt \(\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a}\) =A
A\(=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a+1}}{\sqrt{a}\left(\sqrt{a}+1\right)}\right].\dfrac{a}{\sqrt{a}+1}\)
A\(=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right).\dfrac{a}{\sqrt{a}+1}\)
A\(=\left(\dfrac{a-1}{\sqrt{a}}\right).\dfrac{a}{\sqrt{a}+1}\)
A\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}.\dfrac{a}{\left(\sqrt{a}+1\right)}\)
A=\(\sqrt{a}\left(\sqrt{a}-1\right)\)
A\(=a-\sqrt{a}\)
A\(=a-\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}\)
A\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
Ta có: \(\left(\sqrt{a}-\dfrac{1}{2}\right)^2\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A\ge\dfrac{-1}{4}\)
