\(\left(x+\dfrac{3}{5}\right)\cdot\left(2x-6\right)>0\)
\(+,TH1:\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\\2x-6>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{3}{5}\\2x>6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{3}{5}\\x>3\end{matrix}\right.\)
\(\Leftrightarrow x>3\)
\(+,TH2:\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\2x-6< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\2x< 6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x< 3\end{matrix}\right.\)
\(\Leftrightarrow x< -\dfrac{3}{5}\)
#\(Toru\)
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