sửa đề: \(\frac{\sqrt{x}}{\sqrt{x}+1}\)
đkxđ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}+1\ne0\left(lđ\right)\\x+\sqrt{x}\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\left(\sqrt{x}+1\right)\ne0\\x\ne1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
ta có: \(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right).\frac{1}{\sqrt{x}-1}\)
=\(\left[\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right].\frac{1}{\sqrt{x}-1}\)
=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}.\frac{1}{\sqrt{x}-1}\) =\(\frac{1}{\sqrt{x}}\)
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