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Ta có:
$ \begin{array}{l}
\sqrt {x - 1} + x = 7
\Leftrightarrow \sqrt {x - 1} = 7 - x
\Leftrightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
7 - x \ge 0\\
x - 1 = {(7 - x)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
1 \le x \le 7\\
{x^2} - 15x + 50 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
1 \le x \le 7\\
x = 5\,\,\,V\,\,\,x = 10
\end{array} \right. \Leftrightarrow x = 5\,\,\,(*)
\end{array} $
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