2x^2+3x+can 2x^2 +3x+9=33
can cua ca ba so do minh khong biet viet mong cac ban giup do
\[\begin{array}{l}
2{x^2} + 3x + \sqrt {2{x^2} + 3x + 9} = 33\\
t = 2{x^2} + 3x \Rightarrow t + \sqrt {t + 9} = 33\\
t \ge - 9\\
pt \Leftrightarrow t + 9 = {t^2} - 66t + {33^2}\\
\Leftrightarrow {t^2} - 67t + 1080 = 0\\
\Leftrightarrow t = 40 \vee t = 27\\
\Rightarrow \left[ \begin{array}{l}
2{x^2} + 3x = 40\\
2{x^2} + 3x = 27
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 3 \pm \sqrt {329} }}{4}\\
x = 3 \vee x = - \dfrac{9}{2}
\end{array} \right.\\
f(x) = 2{x^2} + 3x + 9\\
\Delta = {3^2} - 4.2.9 = - 63 < 0\\
\Rightarrow f(x) > 0\forall x\\
\Rightarrow S = \left\{ {\frac{{ - 3 \pm \sqrt {329} }}{4};3; - \dfrac{9}{2}} \right\}
\end{array}\]