\[\begin{array}{l}
y = \left( {3 + x} \right){\left( {x - 2} \right)^2}\\
\Rightarrow y' = {\left( {x - 2} \right)^2} + 2\left( {x - 2} \right)\left( {x + 3} \right)\\
\Rightarrow y' = 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} + 2\left( {x - 2} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 2 + 2x + 6} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {3x - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
3x - 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \frac{4}{3}
\end{array} \right.\\
Bang\,\,xet\,\,dau:\\
- \infty \,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{4}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\, + \infty \\
Vay\,\,hs\,\,\,DB\,\,\,tren\,\,\,\left( { - \infty ;\,\,\frac{4}{3}} \right)\,\,\,\,\,va\,\,\,\,\left( {2; + \infty } \right)\\
hs\,\,\,NB\,\,\,tren\,\,\,\,\left( {\frac{4}{3};\,\,2} \right).
\end{array}\]
