Đáp án:
\(\left[ \begin{array}{l}
x = - 3\\
x = 6
\end{array} \right.\)
Giải thích các bước giải:
Đặt
\(\begin{array}{l}
DK:6 \ge x \ge - 3\\
\sqrt {x + 3} + \sqrt {6 - x} = t\left( {t \ge 0} \right)\\
\to x + 3 + 2\sqrt {\left( {x + 3} \right)\left( {6 - x} \right)} + 6 - x = {t^2}\\
\to 2\sqrt {\left( {x + 3} \right)\left( {6 - x} \right)} + 9 = {t^2}\\
\to \sqrt {\left( {x + 3} \right)\left( {6 - x} \right)} = \dfrac{{{t^2} - 9}}{2}\\
Pt \to t - \dfrac{{{t^2} - 9}}{2} = 3\\
\to - {t^2} + 2t + 9 - 6 = 0\\
\to - {t^2} + 2t + 3 = 0\\
\to \left[ \begin{array}{l}
t = 3\\
t = - 1\left( l \right)
\end{array} \right.\\
\to \sqrt {\left( {x + 3} \right)\left( {6 - x} \right)} = \dfrac{{{3^2} - 9}}{2}\\
\to \left( {x + 3} \right)\left( {6 - x} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 3\\
x = 6
\end{array} \right.\left( {TM} \right)
\end{array}\)
