điều kiện xác định : \(x\ge0;x\ne1\)
ta có : \(A=\left(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+x-\sqrt{x}}{\sqrt{x}-1}\right)\) \(\Leftrightarrow A=\left(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\dfrac{x}{\sqrt{x}-1}\right)\)
\(\Leftrightarrow A=\left(\dfrac{x-\sqrt{x}+1-x+1}{\sqrt{x}-1}\right):\left(\dfrac{x}{\sqrt{x}-1}\right)=\dfrac{2-\sqrt{x}}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{x}=\dfrac{2-\sqrt{x}}{x}\)
