\(A=\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{\sqrt{x}}{2+\sqrt{x}}-\dfrac{6+\sqrt{x}}{4-x}\)
\(=\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{\sqrt{x}}{2+\sqrt{x}}-\dfrac{6+\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}\left(2+\sqrt{x}\right)+\sqrt{x}\left(2-\sqrt{x}\right)-6-\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}=\dfrac{3\left(\sqrt{x}-2\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}=-\dfrac{3}{\sqrt{x}+2}\)
b) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\Rightarrow-\dfrac{3}{\sqrt{x}+2}\ge-\dfrac{3}{2}\)
\(\Rightarrow A_{min}=-\dfrac{3}{2}\) khi \(x=0\)
a) \(A=\left(\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{\sqrt{x}}{2+\sqrt{x}}\right)-\dfrac{6+\sqrt{x}}{4-x}\)
\(=\dfrac{\sqrt{x}\left(2+\sqrt{x}\right)+\sqrt{x}\left(2-\sqrt{x}\right)-\left(6+\sqrt{x}\right)}{4-x}\)
\(=\dfrac{2\sqrt{x}+x+2\sqrt{x}-x-6-\sqrt{x}}{4-x}\)
\(=\dfrac{3\sqrt{x}-6}{4-x}=\dfrac{-3\left(2-\sqrt{x}\right)}{4-x}=\dfrac{-3}{\sqrt{x}+2}\)
b) \(A_{min}\Leftrightarrow\dfrac{-3}{\sqrt{x}+2}_{min}\)
Ta có \(x\ge0\) \(\Rightarrow\sqrt{x}\ge0\) \(\Rightarrow\sqrt{x}+2\ge2\) \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\) \(\Rightarrow\dfrac{-3}{\sqrt{x}+2}\ge\dfrac{3}{2}\)
Dấu "=" xáy ra khi \(\sqrt{x}+2=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
Vậy \(A_{min}=\dfrac{3}{2}\) khi \(x=1\)
