Em tách từng ý ra để hỏi nhé.
\(\begin{array}{l}
a)\,\sqrt {x - 2} + \sqrt {10 - x} = {x^2} - 12x + 40\\
ĐK:\,2 \le x \le 10\\
Đặt\,\,\sqrt {x - 2} + \sqrt {10 - x} = t\,\,\left( {t \ge 0} \right)\\
\Rightarrow {t^2} = 8 + 2\sqrt { - {x^2} + 12x - 20} \\
\Rightarrow {x^2} - 12x + 20 = - {\left( {\dfrac{{{t^2} - 8}}{2}} \right)^2}\\
pt \Rightarrow t = 20 - {\left( {\dfrac{{{t^2} - 8}}{2}} \right)^2}\\
\Leftrightarrow 4t = 80 - \left( {{t^4} - 16{t^2} + 64} \right)\\
\Leftrightarrow {t^4} - 16{t^2} - 16 + 4t = 0\\
\Leftrightarrow {t^2}\left( {t - 4} \right)\left( {t + 4} \right) + 4\left( {t - 4} \right) = 0\\
\Leftrightarrow \left( {t - 4} \right)\left( {{t^3} + 4{t^2} + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 4\\
{t^3} + 4{t^2} + 4 = 0\left( {VL\,do\,\,t \ge 0} \right)
\end{array} \right.\\
\Rightarrow \sqrt {x - 2} + \sqrt {10 - x} = 4\\
\Leftrightarrow {x^2} - 12x + 20 = - {\left( {\dfrac{{{4^2} - 8}}{2}} \right)^2}\\
\Leftrightarrow {x^2} - 12x + 36 = 0\\
\Leftrightarrow {\left( {x - 6} \right)^2} = 0\\
\Rightarrow x = 6\left( {tm} \right)
\end{array}\)
