Đáp án: -9/8
Giải thích các bước giải:
$\begin{array}{l}
x\left( {2x - 3} \right)\\
= 2{x^2} - 3x\\
= 2\left( {{x^2} - \frac{3}{2}x} \right)\\
= 2\left( {{x^2} - 2.\frac{3}{4}.x + {{\left( {\frac{3}{4}} \right)}^2}} \right) - 2.{\left( {\frac{3}{4}} \right)^2}\\
= 2{\left( {x - \frac{3}{4}} \right)^2} - 2.\frac{9}{{16}}\\
= 2{\left( {x - \frac{3}{4}} \right)^2} - \frac{9}{8}\\
do{\left( {x - \frac{3}{4}} \right)^2} \ge 0 \Rightarrow 2{\left( {x - \frac{3}{4}} \right)^2} \ge 0\\
\Rightarrow 2{\left( {x - \frac{3}{4}} \right)^2} - \frac{9}{8} \ge \frac{{ - 9}}{8}\forall x\\
\Rightarrow GTNN\,la\,\,\frac{{ - 9}}{8} \Leftrightarrow x = \frac{3}{4}
\end{array}$
