Đáp án:
\(x=1\)
Giải thích các bước giải:
$\begin{array}{l}
DK:\,\,2x - 2 \ge 0 \Leftrightarrow x \ge 1\\
\sqrt {2x - 2} = 1 - x\\
\Leftrightarrow \left\{ \begin{array}{l}
1 - x \ge 0\\
2x - 2 = {\left( {1 - x} \right)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
2x - 2 = 1 - 2x + {x^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
{x^2} - 4x + 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
\left( {x - 1} \right)\left( {x - 3} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
\left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 1\,\,\left( {tmdk} \right)
\end{array}$
Vậy \(x=1\).
