Đáp án:
$\begin{array}{l}
A = 2\sqrt 3 x + 3{x^2} + 1\\
= {\left( {\sqrt 3 x} \right)^2} + 2\sqrt 3 x + 1\\
= {\left( {\sqrt 3 x + 1} \right)^2} \ge 0\\
\Leftrightarrow GTNN:A = 0\,khi:x = - \dfrac{1}{{\sqrt 3 }} = \dfrac{{ - \sqrt 3 }}{3}\\
B = x - \sqrt 2 {x^2} + 1\\
= - \sqrt 2 \left( {{x^2} - \dfrac{1}{{\sqrt 2 }}x} \right) + 1\\
= - \sqrt 2 \left( {{x^2} - 2.x.\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8}} \right) + \sqrt 2 .\dfrac{1}{8} + 1\\
= - \sqrt 2 {\left( {x - \dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{{8 + \sqrt 2 }}{8} \le \dfrac{{8 + \sqrt 2 }}{8}\\
\Leftrightarrow GTLN:B = \dfrac{{8 + \sqrt 2 }}{8}\,khi:x = \dfrac{1}{{2\sqrt 2 }} = \dfrac{{\sqrt 2 }}{8}\\
C = \sqrt 5 x - 2{x^2} + 3\\
= - 2\left( {{x^2} - \dfrac{{\sqrt 5 }}{2}x} \right) + 3\\
= - 2\left( {{x^2} - 2.x.\dfrac{{\sqrt 5 }}{4} + \dfrac{5}{{16}}} \right) + 2.\dfrac{5}{{16}} + 3\\
= - 2{\left( {x - \dfrac{{\sqrt 5 }}{4}} \right)^2} + \dfrac{{29}}{8} \le \dfrac{{29}}{8}\\
GTLN:C = \dfrac{{29}}{8}\,khi:x = \dfrac{{\sqrt 5 }}{4}\\
D = 1 - 3x + \sqrt 2 {x^2}\\
= \sqrt 2 \left( {{x^2} - \dfrac{3}{{\sqrt 2 }}x} \right) + 1\\
= \sqrt 2 \left( {{x^2} - 2.x.\dfrac{3}{{2\sqrt 2 }} + \dfrac{9}{8}} \right) - \sqrt 2 .\dfrac{9}{8} + 1\\
= \sqrt 2 {\left( {x - \dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{{8 - 9\sqrt 2 }}{8} \ge \dfrac{{8 - 9\sqrt 2 }}{8}\\
\Leftrightarrow GTNN:D = \dfrac{{8 - 9\sqrt 2 }}{8}\,khi:x = \dfrac{3}{{2\sqrt 2 }} = \dfrac{{3\sqrt 2 }}{8}
\end{array}$
