Question

5x(x+3)+2(3+x)=0

Tìm x

Answer 1

\(5x\left(x+3\right)+2\left(3+x\right)=0\)
\(\Rightarrow5x\left(x+3\right)+2\left(x+3\right)=0\)
\(\Rightarrow\left(5x+2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x+2=0\\x+3=0\end{matrix}\right.\)            \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2}{5}\\x=-3\end{matrix}\right.\)

                 Vậy \(x\in\left\{\dfrac{-2}{5};-3\right\}\)

Answer 2

Help

Answer 3

\(5x\left(x+3\right)+2\left(3+x\right)=0\)

\(\Leftrightarrow5x^2+15x+6+2x=0\)

\(\Leftrightarrow5x^2+17x+6=0\)

\(\Leftrightarrow\left(5x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\-3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{2}{5};-3\right\}\).

Answer 4

`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`