\(2x-12-x=0\)
\(\Leftrightarrow2x-x=12\Leftrightarrow x=12\)
\(\left(x-7\right)\left(2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\2x=8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)
2x - 12 - x = 0
\(\Leftrightarrow\)2x - x = 12
\(\Leftrightarrow\)x=12
=))
\(\left(x-7\right).\left(2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\2x=8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)
Vậy \(x\in\left\{7;4\right\}\)
( x - 7) . ( 2x - 8) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x-8=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0+7\\2x=8\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}\)
Vậy:.............................