ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ge0\end{matrix}\right.\)
Ta có: \(G=\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{x-1}{\sqrt{x}+1}\)
\(=\frac{\left(x-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}-x\sqrt{x}+x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x-1}{x-1}=1\)
Điều kiện: \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(G=\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{x-1}{\sqrt{x}+1}\) \(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-\sqrt{x}+1\) \(=1\)
