Đáp án:
$\begin{array}{l}
1)f\left( x \right) \ge 0\forall x\\
\Rightarrow 2{x^2} + 2mx + {m^2} - 2 \ge 0\forall x\\
\Rightarrow \Delta ' \le 0\\
\Rightarrow {m^2} - 2.\left( {{m^2} - 2} \right) \le 0\\
\Rightarrow {m^2} - 2m + 4 \le 0\\
\Rightarrow {m^2} \ge 4\\
\Rightarrow \left[ \begin{array}{l}
m \ge 2\\
m \le - 2
\end{array} \right.\\
2)2{x^2} + 2mx + {m^2} - 2 = 0\\
\Delta ' \ge 0\\
\Rightarrow {m^2} \le 4\\
\Rightarrow - 2 \le m \le 2\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - m\\
{x_1}{x_2} = \dfrac{{{m^2} - 2}}{2}
\end{array} \right.\\
A = \left| {2{x_1}{x_2} + {x_1} + {x_2} - 4} \right|\\
= \left| {{m^2} - 2 - m - 4} \right|\\
= \left| {{m^2} - m - 6} \right|\\
= \left| {{{\left( {m - \dfrac{1}{2}} \right)}^2} - \dfrac{{25}}{4}} \right|\\
Do:{\left( {m - \dfrac{1}{2}} \right)^2} - \dfrac{{25}}{4} \ge - \dfrac{{25}}{4}\\
\Rightarrow \left| {{{\left( {m - \dfrac{1}{2}} \right)}^2} - \dfrac{{25}}{4}} \right| \le \dfrac{{25}}{4}\\
\Rightarrow \max A = \dfrac{{25}}{4} \Leftrightarrow m = \dfrac{1}{2}
\end{array}$
Bài 3:
$\begin{array}{l}
\sin a = \dfrac{4}{5}\\
Do:0 < a < \dfrac{\pi }{2}\\
\Rightarrow \cos a > 0\\
\Rightarrow \cos a = \sqrt {1 - {{\sin }^2}a} = \sqrt {1 - \dfrac{{16}}{{25}}} = \dfrac{3}{5}\\
\Rightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} = \dfrac{4}{5}\\
\Rightarrow P = 5.\cos a + 3\tan a\\
= 5.\dfrac{3}{5} + 3.\dfrac{4}{5} = \dfrac{{27}}{5}
\end{array}$
