Đáp án:
$\begin{array}{l}
1)y = \left( {{x^2} + 3x} \right)\sqrt {4 - {x^2}} \\
\Leftrightarrow y' = \left( {2x + 3} \right).\sqrt {4 - {x^2}} + \left( {{x^2} + 3x} \right).\dfrac{{ - 2x}}{{2\sqrt {4 - {x^2}} }}\\
= \left( {2x + 3} \right).\sqrt {4 - {x^2}} - \dfrac{{x\left( {{x^2} + 3x} \right)}}{{\sqrt {4 - {x^2}} }}\\
= \dfrac{{\left( {2x + 3} \right)\left( {4 - {x^2}} \right) - {x^3} - 3{x^2}}}{{\sqrt {4 - {x^2}} }}\\
= \dfrac{{8x - 2{x^3} + 12 - 3{x^2} - {x^3} - 3{x^2}}}{{\sqrt {4 - {x^2}} }}\\
= \dfrac{{ - 3{x^3} - 6{x^2} + 8x + 12}}{{\sqrt {4 - {x^2}} }}\\
2)y = \sin 3x + \cos 6x\\
\Leftrightarrow y' = 3.\cos 3x - 6\sin 6x\\
3)y = x.\sin x + \cos x\\
\Leftrightarrow y' = \sin x + x.\cos x - \sin x\\
= x.\cos x\\
3)y = {x^3} - 5{x^2} + 7x + 2020\\
\Leftrightarrow y' = 3{x^2} - 10x + 7\\
Khi:y' \le 0\\
\Leftrightarrow 3{x^2} - 10x + 7 \le 0\\
\Leftrightarrow \left( {3x - 7} \right)\left( {x - 1} \right) \le 0\\
\Leftrightarrow 1 \le x \le \dfrac{7}{3}\\
Vậy\,1 \le x \le \dfrac{7}{3}
\end{array}$
