a) điều kiện xác định : \(x\ge0;x\ne1\)
b) ta có : \(Q=\dfrac{1}{2\sqrt{x}-2}+\dfrac{1}{2\sqrt{x}+2}+\dfrac{x}{1-x}\)
\(\Leftrightarrow Q=\dfrac{1}{2\left(\sqrt{x}-1\right)}+\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow Q=\dfrac{\sqrt{x}+1+\sqrt{x}-1-2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow Q=\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\)
c) thay \(x=\dfrac{4}{9}\) vào \(Q\) ta có \(Q=\dfrac{-2\sqrt{\dfrac{4}{9}}}{\sqrt{\dfrac{4}{9}}+1}=\dfrac{-4}{5}\)
d) để \(A=\dfrac{-1}{2}\Leftrightarrow\dfrac{-2\sqrt{x}}{\sqrt{x}+1}=\dfrac{-1}{2}\Leftrightarrow-\sqrt{x}-1=-4\sqrt{x}\)
\(\Leftrightarrow3\sqrt{x}=1\Leftrightarrow\sqrt{x}=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{9}\)
e) ta có \(Q=\dfrac{-2\sqrt{x}}{\sqrt{x}+1}=\dfrac{-2\sqrt{x}-2+2}{\sqrt{x}+1}=-2+\dfrac{2}{\sqrt{x}+1}\)
\(\Rightarrow\) \(Q\) nguyên \(\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\) nguyên \(\Leftrightarrow\sqrt{x}+1\) thuộc ước của \(2\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\in\left\{\pm1;\pm2\right\}\) \(\Leftrightarrow x\in\left\{0;1\right\}\)
