Đáp án:
$\begin{array}{l}
1)y = \left( {m + 1} \right){x^3} + m{x^2} + 4x - 3\\
\Leftrightarrow y' = 3\left( {m + 1} \right).{x^2} + 2mx + 4\\
y' < 0\forall x\\
\Leftrightarrow 3\left( {m + 1} \right).{x^2} + 2mx + 4 < 0\forall x\\
\Leftrightarrow \left\{ \begin{array}{l}
3\left( {m + 1} \right) < 0\\
\Delta ' < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - 1\\
{m^2} - 3\left( {m + 1} \right).4 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - 1\\
{m^2} - 12m - 12 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - 1\\
6 - 4\sqrt 3 < m < 6 + 4\sqrt 3
\end{array} \right.\\
\Leftrightarrow 6 - 4\sqrt 3 < m < - 1\\
2)y = \sin x - \cot \left( {{x^2} - \dfrac{\pi }{3}} \right)\\
\Leftrightarrow y' = \cos x + 2x.\dfrac{1}{{{{\sin }^2}\left( {{x^2} - \dfrac{\pi }{3}} \right)}}\\
3)y = \dfrac{{{{\left( {5{x^2} - 3x} \right)}^3}}}{{\cos x}}\\
y' = \dfrac{{3.\left( {10x - 3} \right).{{\left( {5{x^2} - 3x} \right)}^2}.\cos x + \sin x.{{\left( {5{x^2} - 3x} \right)}^3}}}{{{{\cos }^2}x}}
\end{array}$
