Đáp án:
\(x=6.\)
Giải thích các bước giải:
\[\begin{array}{l}
\sqrt {x - 2} + \sqrt {10 - x} = {x^2} - 12x + 40\,\,\left( * \right)\\
DK:\,\,\,2 \le x \le 10\\
Dat\,\,\,t = \sqrt {x - 2} + \sqrt {10 - x} > 0\\
\Rightarrow {t^2} = x - 2 + 10 - x + 2\sqrt {\left( {x - 2} \right)\left( {10 - x} \right)} = 8 + 2\sqrt {12x - {x^2} - 20} \\
\Rightarrow \sqrt {12x - {x^2} - 20} = \frac{{{t^2} - 8}}{2}\\
\Leftrightarrow 12x - {x^2} - 20 = {\left( {\frac{{{t^2} - 8}}{2}} \right)^2}\\
\Leftrightarrow {x^2} - 12x = - 20 - {\left( {\frac{{{t^2} - 8}}{2}} \right)^2}\\
\Rightarrow \left( * \right) \Leftrightarrow t = - 20 - {\left( {\frac{{{t^2} - 8}}{2}} \right)^2} + 40\\
\Leftrightarrow t = 20 - \frac{{{t^4} - 16{t^2} + 64}}{4}\\
\Leftrightarrow 4t = 80 - {t^4} + 16{t^2} - 64\\
\Leftrightarrow {t^4} - 16{t^2} + 4t - 16 = 0\\
\Leftrightarrow \left( {t - 4} \right)\left( {{t^3} + 4{t^2} + 4} \right) = 0\\
\Leftrightarrow t = 4\,\,\,\,\left( {do\,\,t > 0 \Rightarrow {t^3} + 4{t^2} + 4 = 0\,\,\,VN} \right)\\
\Leftrightarrow 8 + 2\sqrt {12x - {x^2} - 20} = 16\\
\Leftrightarrow 2\sqrt {12x - {x^2} - 20} = 8\\
\Leftrightarrow \sqrt {12x - {x^2} - 20} = 4\\
\Leftrightarrow 12x - {x^2} - 20 = 16\\
\Leftrightarrow {x^2} - 12x + 36 = 0\\
\Leftrightarrow x = 6\,\,\,\left( {tm} \right) \ge
\end{array}\]
