Đáp án:
\(x=2.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {2x - 3} + \sqrt {5 - 2x} - {x^2} + 4x - 6 = 0\,\,\,\,\,\left( * \right)\\
DK:\,\,\,\frac{3}{2} \le x \le \frac{5}{2}\\
Dat\,\,t = \sqrt {2x - 3} + \sqrt {5 - 2x} \,\,\left( {t \ge 0} \right)\\
\Rightarrow {t^2} = 2x - 3 + 5 - 2x + 2\sqrt {\left( {2x - 3} \right)\left( {5 - 2x} \right)} \\
\Leftrightarrow {t^2} = 2 + 2\sqrt { - 4{x^2} + 16x - 15} \\
\Leftrightarrow {t^2} - 2 = 2\sqrt { - 4\left( {{x^2} - 4x} \right) - 15} \\
\Leftrightarrow {\left( {{t^2} - 2} \right)^2} = 4\left[ { - 4\left( {{x^2} - 4x} \right) - 15} \right]\\
\Leftrightarrow {t^4} - 4{t^2} + 4 = - 16\left( {{x^2} - 4x} \right) - 60\\
\Leftrightarrow - \left( {{x^2} - 4x} \right) = \frac{{{t^4} - 4{t^2} + 64}}{{16}}\\
\Rightarrow \left( * \right) \Leftrightarrow t + \frac{{{t^4} - 4{t^2} + 64}}{{16}} - 6 = 0\\
\Leftrightarrow 16t + {t^4} - 4{t^2} + 64 - 96 = 0\\
\Leftrightarrow {t^4} - 4{t^2} + 16t - 32 = 0\\
\Leftrightarrow {t^2}\left( {{t^2} - 4} \right) + 16\left( {t - 2} \right) = 0\\
\Leftrightarrow {t^2}\left( {t - 2} \right)\left( {t + 2} \right) + 16\left( {t - 2} \right) = 0\\
\Leftrightarrow \left( {t - 2} \right)\left[ {{t^2}\left( {t + 2} \right) + 16} \right] = 0\\
\Leftrightarrow t - 2 = 0\,\,\,\,\,\left( {do\,\,\,{t^2}\left( {t + 2} \right) + 16 > 0\,\,\,\forall t \ge 0} \right)\\
\Leftrightarrow t = 2\,\,\,\left( {tm} \right)\\
\Leftrightarrow - \left( {{x^2} - 4x} \right) = \frac{{{t^4} - 4{t^2} + 64}}{{16}} = 4\\
\Leftrightarrow - {x^2} + 4x - 4 = 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} = 0\\
\Leftrightarrow x - 2 = 0\\
\Leftrightarrow x = 2\,\,\,\left( {tm} \right).
\end{array}\)
