Xét tính đơn điệu của hàm số sau Y=(3+x)(x-2)^2

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