Đáp án:
$\begin{array}{l}
D = 5{x^2} + 3x + 18\\
= 5\left( {{x^2} + \dfrac{3}{5}x} \right) + 18\\
= 5.\left( {{x^2} + 2.x.\dfrac{3}{{10}} + \dfrac{9}{{100}} - \dfrac{9}{{100}}} \right) + 18\\
= 5.{\left( {x + \dfrac{3}{{10}}} \right)^2} - 5.\dfrac{9}{{100}} + 18\\
= 5.{\left( {x + \dfrac{3}{{10}}} \right)^2} + \dfrac{{351}}{{20}}\\
Do:5.{\left( {x + \dfrac{3}{{10}}} \right)^2} \ge 0\\
\Rightarrow 5.{\left( {x + \dfrac{3}{{10}}} \right)^2} + \dfrac{{351}}{{20}} \ge \dfrac{{351}}{{20}}\\
\Rightarrow GTNN:D = \dfrac{{351}}{{20}} \Leftrightarrow x = - \dfrac{3}{{10}}\\
E = {\left( {x + 1} \right)^2} + {\left( {x + 3} \right)^2} + 2\left( {x - 5} \right)\\
= 2{x^2} + 10x\\
= 2\left( {{x^2} + 5x} \right)\\
= 2\left( {{x^2} + 2.x.\dfrac{5}{2} + \dfrac{{25}}{4}} \right) - 2.\dfrac{{25}}{4}\\
= 2.{\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{2} \ge - \dfrac{{25}}{2}\\
\Rightarrow GTNN:E = - \dfrac{{25}}{2} \Leftrightarrow x = - \dfrac{5}{2}\\
F = {\left( {x + 4} \right)^2} + {\left( {2x + 1} \right)^2}\\
= {x^2} + 8x + 16 + 4{x^2} + 4x + 1\\
= 5{x^2} + 12x + 17\\
= 5.\left( {{x^2} + \dfrac{{12}}{5}x} \right) + 17\\
= 5.\left( {{x^2} + 2.x.\dfrac{6}{5} + \dfrac{{36}}{{25}}} \right) - 5.\dfrac{{36}}{{25}} + 17\\
= 5.{\left( {x + \dfrac{6}{5}} \right)^2} + \dfrac{{49}}{5} \ge \dfrac{{49}}{5}\\
\Rightarrow GTNN:F = \dfrac{{49}}{5} \Leftrightarrow x = - \dfrac{6}{5}
\end{array}$
