\(35-\left[\left(2x-3\right)^2:7\right]=28\)
\(\Rightarrow\left[\left(2x-3\right)^2:7\right]=35-28\)
\(\Rightarrow\left(2x-3\right)^2:7=7\)
\(\Rightarrow\left(2x-3\right)^2=1\)
\(\Rightarrow2x-3=\pm1\)
\(\Rightarrow x=2\) hay \(x=1\)
35 - [(2\(x\) - 3)2:7 ] = 28
(2\(x-3\))2 : 7 = 35 - 28
(2\(x\) - 3)2 : 7 = 7
(2\(x\) - 3)2 = 7 \(\times\) 7
(2\(x-3\))2 = 72
\(\left[{}\begin{matrix}2x-3=-7\\2x-3=7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-7+3\\2x=7+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Vậy \(x\in\) {-2; 5}
