Đáp án: x=2
Giải thích các bước giải:
$\begin{array}{l}
\sqrt {x + 2} - \sqrt {x - 1} = \sqrt {2x - 3} \left( 1 \right)\\
Đkxđ:\left\{ \begin{array}{l}
x + 2 \ge 0\\
x - 1 \ge 0\\
2x - 3 \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge - 2\\
x \ge 1\\
x \ge \frac{3}{2}
\end{array} \right. \Rightarrow x \ge \frac{3}{2}\\
\left( 1 \right) \Rightarrow \sqrt {x + 2} = \sqrt {x - 1} + \sqrt {2x - 3} \\
\Rightarrow x + 2 = x - 1 + 2\sqrt {x - 1} .\sqrt {2x - 3} + 2x - 3\\
\Rightarrow - 2x + 6 = 2\sqrt {\left( {x - 1} \right)\left( {2x - 3} \right)} \\
\Rightarrow 3 - x = \sqrt {2{x^2} - 5x + 3} \left( {dk:x \le 3} \right)\\
\Rightarrow {x^2} - 6x + 9 = 2{x^2} - 5x + 3\\
\Rightarrow {x^2} + x - 6 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x = - 3\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 2
\end{array}$
