Question

Tập hợp các số nguyên để là

Answer 1

\(\left(x+3\right)\left(6+2x\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot2\left(3+x\right)=0\)

\(\Leftrightarrow2\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

Vậy..........................

Answer 2

Ta có:

\(\left(x+3\right)\left(6+2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0-3\\2x=0-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-3\end{matrix}\right.\)

Vậy \(x=\left\{-3\right\}\)