\(5x\left(x+3\right)+2\left(3+x\right)=0\)
\(\Rightarrow5x\left(x+3\right)+2\left(x+3\right)=0\)
\(\Rightarrow\left(5x+2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x+2=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2}{5}\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-2}{5};-3\right\}\)
\(5x\left(x+3\right)+2\left(3+x\right)=0\)
\(\Leftrightarrow5x^2+15x+6+2x=0\)
\(\Leftrightarrow5x^2+17x+6=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\-3\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{2}{5};-3\right\}\).
`5x(x + 3) + 2(3 + x) = 0`
`<=> 5x(x + 3) + 2(x + 3) = 0`
`<=> (x + 3)(5x + 2) = 0`
`<=>` $\left[\begin{matrix} x + 3 = 0\\ 5x + 2 = 0\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x = -3\\ x = \dfrac{-2}{5}\end{matrix}\right.$
Vậy `x = -3` hoặc `x = (-2)/5`