Giải thích các bước giải:
a,
ĐKXĐ: \(\frac{3}{2} \le x \le \frac{5}{2}\)
Ta có:
\(\begin{array}{l}
\sqrt {2x - 3} + \sqrt {5 - 2x} = 2{x^2} - 3x\\
\Leftrightarrow \left( {\sqrt {2x - 3} - 1} \right) + \left( {\sqrt {5 - 2x} - 1} \right) = 2{x^2} - 3x - 2\\
\Leftrightarrow \frac{{2x - 3 - 1}}{{\sqrt {2x - 3} + 1}} + \frac{{5 - 2x - 1}}{{\sqrt {5 - 2x} + 1}} = \left( {x - 2} \right)\left( {2x + 1} \right)\\
\Leftrightarrow \frac{{2\left( {x - 2} \right)}}{{\sqrt {2x - 3} + 1}} - \frac{{2\left( {x - 2} \right)}}{{\sqrt {5 - 2x} + 1}} = \left( {x - 2} \right)\left( {2x + 1} \right)\\
\Leftrightarrow \left( {x - 2} \right)\left[ {\left( {2x + 1} \right) + \frac{2}{{\sqrt {5 - 2x} + 1}} - \frac{2}{{\sqrt {2x - 3} + 1}}} \right] = 0\\
\frac{3}{2} \le x \le \frac{5}{2} \Rightarrow \left\{ \begin{array}{l}
\sqrt {2x - 3} + 1 \ge 1 \Rightarrow \frac{2}{{\sqrt {2x - 3} + 1}} \le \frac{2}{1} = 2\\
2x + 1 \ge 2.\frac{3}{2} + 1 = 4
\end{array} \right.\\
\Rightarrow \left( {2x + 1} \right) + \frac{2}{{\sqrt {5 - 2x} + 1}} - \frac{2}{{\sqrt {2x - 3} + 1}} > 0,\,\,\,\,\forall x \in \left[ {\frac{3}{2};\,\frac{5}{2}} \right]\\
\Rightarrow x - 2 = 0\\
\Leftrightarrow x = 2\left( {t/m} \right)
\end{array}\)
Vậy \(x = 2\)
b,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge \frac{2}{3}\\
y \ge - 2
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\sqrt {3x - 2} + 4\sqrt {y + 2} = 14\\
3\sqrt {3x - 2} - \sqrt {y + 2} = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {3x - 2} = 14 - 4\sqrt {y + 2} \\
3.\left( {14 - 4\sqrt {y + 2} } \right) - \sqrt {y + 2} = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {3x - 2} = 14 - 4\sqrt {y + 2} \\
42 - 12\sqrt {y + 2} - \sqrt {y + 2} = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {y + 2} = 3\\
\sqrt {3x - 2} = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 7
\end{array} \right.\left( {t/m} \right)
\end{array}\)
Vậy \(\left\{ \begin{array}{l}
x = 2\\
y = 7
\end{array} \right.\)
