Đáp án:
$\begin{array}{l}
31)\\
A = {\left( {3x - 2} \right)^2} + {\left( {3x + 2} \right)^2} + 2\left( {9{x^2} - 6} \right)\\
= {\left( {3x - 2} \right)^2} + {\left( {3x + 2} \right)^2} + 2\left( {9{x^2} - 1} \right) - 4\\
= {\left( {3x - 2} \right)^2} + 2.\left( {3x - 2} \right)\left( {3x + 2} \right) + {\left( {3x + 2} \right)^2} - 4\\
= {\left( {3x - 2 + 3x + 2} \right)^2} - 4\\
= {\left( {6x} \right)^2} - 4\\
= 36{x^2} - 4\\
Khi:x = \dfrac{{ - 1}}{3}\\
\Leftrightarrow A = 36.{\left( { - \dfrac{1}{3}} \right)^2} - 4 = 4 - 4 = 0\\
\Leftrightarrow D\\
32)\\
K = \left( {{x^2} + 2x + 3} \right)\left( {{x^2} + 2x + 4} \right)\\
Dat:{x^2} + 2x + 1 = a = {\left( {x + 1} \right)^2} \ge 0\\
K = \left( {a + 2} \right)\left( {a + 3} \right)\\
= {a^2} + 5a + 6 \ge 6\left( {do:a \ge 0} \right)\\
\Leftrightarrow K \ge 6\\
\Leftrightarrow GTNN:K = 6
\end{array}$
=> Dấu = xảy ra khi x=-1
=> ko có đáp án đúng
