đkxđ: x≥0; x≠4
\(A=\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right):\dfrac{\sqrt{x}+3}{2\sqrt{x}-x}=\left[\dfrac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+3}=\dfrac{\left(2+\sqrt{x}-2+\sqrt{x}\right)\left(2+\sqrt{x}+2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+3}=\dfrac{2\sqrt{x}\cdot4}{2+\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+3}=\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
Ta có: \(A>0\Leftrightarrow\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}>0\)
Ta thấy: \(\sqrt{x}+2>0\forall x\ge0;\sqrt{x}+3>0\forall x\ge0\)
\(\Rightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)>0\)
⇒ Để A > 0 thì 8x > 0 <=> x>0
Vậy x>0 thì A>0
