a) ĐK: \(x\ge0;x\ne1\)
Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
<=> \(A=\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right]:\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right]:\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2}{\left(\sqrt{x}+1\right)^2}\)
b) Để A nguyên => \(\dfrac{2}{\left(\sqrt{x}+1\right)^2}\) nguyên
=> \(2⋮\left(\sqrt{x}+1\right)^2\) => \(\left(\sqrt{x}+1\right)^2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Mà \(\left(\sqrt{x}+1\right)^2\ge1\) => \(\left(\sqrt{x}+1\right)^2\in\left\{1;2\right\}\)
=> \(\left[{}\begin{matrix}\sqrt{x}+1=1\\\sqrt{x}+1=\sqrt{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=0\\x=3-2\sqrt{2}\end{matrix}\right.\) (TM)
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