`(2x - 1) xx 2 = 25`
`=> 2x - 1 = 25/2 = 12,5`
`=> 2x = 13,5`
`=> x = (13,5)/2`
`=> x = 6,75`.
\(\left(2x-1\right)^2=25\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
`(2x-1)^2 = 25`
`=>(2x-1)^2 = (+-5)^2`
`TH1:2x-1=5`
`=>2x=6`
`=>x=3`
`TH2:2x-1=-5`
`=>2x=-4`
`=>x=-2`
Vậy `x in {-2;3}`
`(2x - 1)^2 = 25`
`=> (2x - 1)^2 = 5^2`
`=>`\(\left\{{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
`=>`\(\left\{{}\begin{matrix}2x=5+1\\2x=-5+1\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}2x=6\\2x=4\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}x=6:2\\x=4:2\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)