Bài 2
a) Để P có nghĩa
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
vậy để P có nghĩ thì \(x\ge0;x\ne1\)
b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{-2\sqrt{x}}{\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}\right).\dfrac{\sqrt{x}-1}{-2\sqrt{x}}\)
\(=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{x-1}.\dfrac{\sqrt{x}-1}{-2\sqrt{x}}\)
\(=\dfrac{-2\sqrt{x}}{x-1}.\dfrac{\sqrt{x}-1}{-2\sqrt{x}}\)
\(\dfrac{1}{\sqrt{x}+1}\)