Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {2x + 5} = \sqrt {1 - x} \\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x + 5 \ge 0\\
1 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - \frac{5}{2}\\
x \le 1
\end{array} \right. \Leftrightarrow - \frac{5}{2} \le x \le 1\\
\sqrt {2x + 5} = \sqrt {1 - x} \\
\Leftrightarrow 2x + 5 = 1 - x\\
\Leftrightarrow 2x + x = 1 - 5\\
\Leftrightarrow 3x = - 4\\
\Leftrightarrow x = - \frac{4}{3}\,\,\,\,\,\,\,\,\,\,\left( {t/m} \right)\\
b,\\
DKXD:\,\,\,\,\,\left\{ \begin{array}{l}
{x^2} - x \ge 0\\
3 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x\left( {x - 1} \right) \ge 0\\
x \le 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 1\\
x \le 0
\end{array} \right.\\
x \le 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \le 0\\
1 \le x \le 3
\end{array} \right.\\
\sqrt {{x^2} - x} = \sqrt {3 - x} \\
\Leftrightarrow {x^2} - x = 3 - x\\
\Leftrightarrow {x^2} = 3\\
\Leftrightarrow x = \pm \sqrt 3 \,\,\,\,\,\,\,\,\,\left( {t/m} \right)\\
c,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2{x^2} - 3 \ge 0\\
4x - 3 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \sqrt {\frac{3}{2}} \\
x \le - \sqrt {\frac{3}{2}}
\end{array} \right.\\
x \ge \frac{3}{2}
\end{array} \right. \Leftrightarrow x \ge \frac{{\sqrt 6 }}{2}\\
\sqrt {2{x^2} - 3} = \sqrt {4x - 3} \\
\Leftrightarrow 2{x^2} - 3 = 4x - 3\\
\Leftrightarrow 2{x^2} - 4x = 0\\
\Leftrightarrow 2x\left( {x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\,\,\,\,\,\,\,\,\,\,\left( L \right)\\
x = 2\,\,\,\,\,\,\,\,\,\,\,\,\left( {t/m} \right)
\end{array} \right.
\end{array}\)
