a) điều kiện xác định : \(x>0;x\ne1\)
ta có : \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(\Leftrightarrow Q=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(\Leftrightarrow Q=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(\Leftrightarrow Q=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2}{x-1}\)
d) ta có : \(Q\in Z\) \(\Leftrightarrow\dfrac{2}{x-1}\in Z\Leftrightarrow x-1\) là ước của \(2\)
\(\Leftrightarrow\left(x-1\right)\in\left\{\pm1;\pm2\right\}\) \(\Leftrightarrow x\in\left\{-1;0;2;3\right\}\) \(\Rightarrow x=3\)
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