`(5x+1).(2x-5)=0`
`@TH1:`
`5x+1=0`
`5x=0-1`
`5x=-1`
`x=-1:5`
`x=-1/5`
`@TH2:`
`2x-5=0`
`2x=0+5`
`2x=5`
`x=5:2`
`x=5/2`
Vậy `x= {5/2; (-1)/5}`
`#LeMichael`
\((5x+1)(2x-5)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\2x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=2,5\end{matrix}\right.\)