Ta có:
\(\left(2x-3\right)\left(6-2x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=3\end{matrix}\right.\)
(2x - 3)(6 - 2x) = 0 => 2x-3=0 hoặc 6-2x=0
+) 2x - 3 = 0
2x = 3
x = 3: 2 (loại)
+)6 - 2x = 0
2x = 6
x = 6: 2
x = 3
Vậy x = 3
\(\left(2x-3\right)\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)