\(\left(x-3\right)\left(2x+8\right)\ge0\)
Th1: \(\hept{\begin{cases}x-3\ge0\\2x+8\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge3\\x\ge-4\end{cases}\Rightarrow}x\ge3\)
Th2: \(\hept{\begin{cases}x-3\le0\\2x+8\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\le3\\x\le-4\end{cases}\Rightarrow}x\le-4\)
\(\left(x-3\right).\left(2x+8\right)\ge0\)
\(\Rightarrow\hept{\begin{cases}x-3\ge0\\2x+8\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x\ge3\\2x\ge-8\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x\ge3\\x\ge-4\end{cases}}\)
\(\Rightarrow x\ge3\)
Vậy \(x\ge3\)
Bài giải
\(\left(x-3\right)\left(2x+8\right)\ge0\)
\(TH1\text{ : }\hept{\begin{cases}x-3\le0\\2x+8\le0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x\le3\\2x\le-8\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x\le3\\x\le-4\end{cases}}\) \(\Rightarrow\text{ }x\le3^{\left(1\right)}\)
\(TH2\text{ : }\hept{\begin{cases}x-3\ge0\\2x+8\ge0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x\ge3\\2x\ge-8\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x\ge3\\x\ge-4\end{cases}}\) \(\Rightarrow\text{ }x\ge-4^{\left(2\right)}\)
Từ ( 1 ) và ( 2 ) \(\Rightarrow\text{ }-4\le x\le3\)
\(\Rightarrow\text{ }x\in\left\{-4\text{ ; }-3\text{ ; }-2\text{ ; }-1\text{ ; }0\text{ ; }1\text{ ; }2\text{ ; }3\right\}\)
\(\left(x-3\right)\left(2x+8\right)\ge0\)
TH 1 : \(\hept{\begin{cases}x-3\le0\\2x+8\le0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x\le3\\2x\le-8\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x\le3\\x\le-4\end{cases}}\) \(\Rightarrow\text{ }x\le3^{\left(1\right)}\)
TH 2 : \(\hept{\begin{cases}x-3\ge0\\2x+8\ge0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x\ge3\\2x\ge-8\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x\ge3\\x\ge-4\end{cases}}\) \(\Rightarrow\text{ }x\ge-4^{\left(2\right)}\)
Từ ( 1 ) và ( 2 ) \(\Rightarrow\text{ }-4\le x\le3\)
\(\Rightarrow\text{ }x\in\left\{-4\text{ ; }-3\text{ ; }-2\text{ ; }-1\text{ ; }0\text{ ; }1\text{ ; }2\text{ ; }3\right\}\)
