a) \(\left(x-5\right)\left(2x-3^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\2x=9\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{9}{2}\end{matrix}\right.\)
b) \(2\left(3x-15\right)\left(5-x\right)=0\)
\(\Rightarrow6\left(x-5\right)\left(5-x\right)=0\Rightarrow x=5\)
(x - 5)(2x - 32) = 0
=> \(\left[\begin{array}{} x - 5 = 0\\ 2x - 3^{2} = 0 \end{array} \right.\)=> \(\left[\begin{array}{} x = 0 - 5 = -5\\ 2x = 0 - 3^{2} = 0 - 9 = -9 => x = \dfrac{9}{2} \end{array} \right.\)
b) 2(3x - 15)(5 - x) = 0
(6x - 30)(5 - x) = 0
=> \(\begin{cases} 6x - 30 = 0\\ 5 - x =0 \end{cases} \) => \(\begin{cases} 6x = 0 + 30 = 30 => x = 5\\ x = 0 + 5 = 5 \end{cases} \)
