$a.2x(x-5)-4(5-x)=0$
$⇔2x(x-5)+4(x-5)=0$
$⇔(x-5)(2x+4)=0$
⇔\(\left[ \begin{array}{l}x-5=0\\2x+4=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)
$b.|x-3|=-3x+17$
⇔\(\left[ \begin{array}{l}x-3=-3x+17\\x-3=3x-17\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x+3x=17+3\\x-3x=-17+3\end{array} \right.\)
⇔\(\left[ \begin{array}{l}4x=20\\-2x=-14\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\)
