Question

Tìm x để :

\(\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

Answer 1

\(\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(ĐKXĐ:x\ge0\right)\)

\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-1\ge0\\\sqrt{x}+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}+1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}0\le x\le1\\x\in\varnothing\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

Vậy .................