Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
f\left( x \right) = a{x^2} + bx + c \ge 0,\,\,\,\forall x \in R \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
\Delta \le 0
\end{array} \right.\\
{x^2} - \left( {3m + 1} \right)x + 2{m^2} + 2m \ge 0,\,\,\,\,\forall x \in R\\
\Leftrightarrow \Delta \le 0\\
\Leftrightarrow {\left( {3m + 1} \right)^2} - 4.1.\left( {2{m^2} + 2m} \right) \le 0\\
\Leftrightarrow \left( {9{m^2} + 6m + 1} \right) - \left( {8{m^2} + 8m} \right) \le 0\\
\Leftrightarrow {m^2} - 2m + 1 \le 0\\
\Leftrightarrow {\left( {m - 1} \right)^2} \le 0\\
{\left( {m - 1} \right)^2} \ge 0,\,\,\,\forall m \Rightarrow {\left( {m - 1} \right)^2} = 0 \Leftrightarrow m = 1\\
4,\\
0 < x < \dfrac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin x > 0\\
\cos x > 0
\end{array} \right.\\
{\sin ^2}x + {\cos ^2}x = 1\\
\sin x > 0 \Rightarrow \sin x = \sqrt {1 - {{\cos }^2}x} = \dfrac{4}{5}\\
\sin 2x = 2.\sin x.\cos x = 2.\dfrac{4}{5}.\dfrac{3}{5} = \dfrac{{24}}{{25}}\\
\cos 2x = 2{\cos ^2}x - 1 = 2.{\left( {\dfrac{3}{5}} \right)^2} - 1 = \dfrac{{ - 7}}{{25}}
\end{array}\)
