Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
P = 2{x^2} + 5x + 8 = 2\left( {{x^2} + \frac{5}{2}x + 4} \right)\\
= 2\left( {{x^2} + 2.x.\frac{5}{4} + \frac{{25}}{{16}}} \right) + \frac{{39}}{8}\\
= 2{\left( {x + \frac{5}{4}} \right)^2} + \frac{{39}}{8} \ge \frac{{39}}{8}\\
\Rightarrow {P_{\min }} = \frac{{39}}{8} \Leftrightarrow x = - \frac{5}{4}\\
2,\\
Q = - 3{x^2} + 7x + 15 = - 3.\left( {{x^2} - \frac{7}{3}x - 5} \right)\\
= - 3\left( {{x^2} - 2.x.\frac{7}{6} + \frac{{49}}{{36}}} \right) + \frac{{229}}{{12}}\\
= \frac{{229}}{{12}} - 3{\left( {x - \frac{7}{6}} \right)^2} \le \frac{{229}}{{12}}\\
\Rightarrow {Q_{\max }} = \frac{{229}}{{12}} \Leftrightarrow x = \frac{7}{6}\\
3,\\
{x^2} + 2x + 3 = \left( {{x^2} + 2x + 1} \right) + 2 = {\left( {x + 1} \right)^2} + 2 \ge 2\\
\Rightarrow P = \frac{5}{{{x^2} + 2x + 3}} \le \frac{5}{2}\\
\Rightarrow {P_{\max }} = \frac{5}{2} \Leftrightarrow x = - 1\\
4,\\
- {x^2} + 2x + 5 = - \left( {{x^2} - 2x + 1} \right) + 6 = 6 - {\left( {x - 1} \right)^2} \le 6\\
\Rightarrow P = \frac{6}{{ - {x^2} + 2x + 5}} \ge \frac{6}{6} = 1\\
\Rightarrow {P_{\min }} = 1 \Leftrightarrow x = 1
\end{array}\)
