\(\left(2x-3\right)^2=81\)
\(\Leftrightarrow\left(2x-3\right)^2=\pm\sqrt{81}\)
\(\Leftrightarrow\left(2x-3\right)=\pm9\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=9\\2x-3=-9\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}2x=12\\2x=-6\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=-3\end{array}\right.\)
Vậy \(x\in\left\{6;-3\right\}\)
(2x- 3)2= 81
(2x- 3)2=92
2x-3= 9 2x-3= -9
2x = 9+3 2x = (-9)+3
2x = 12 2x = -6
x = 12:2 x = -6:2
x = 6 x = -3
VAY x = 6 va x = -3
