Giúp em 3 câu này ạ. Em đang cần gấp.

Giải thích các bước giải:

Ta có:

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