$a) (4x-1)(x-3)-(x-3)(5x+2) = 0$
$⇔(x-3)(4x-1-5x-2)=0$
$⇔(x-3)(-x-3)=0$
⇔\(\left[ \begin{array}{l}x-3=0\\-x-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=3\\x=-3\end{array} \right.\)
$Vậy$ $x=3$ $hoặc$ $x=-3$
$b/$
$(x+6)(3x-1)+(x²-36)=0$
$⇔(x+6)(3x-1)+(x-6)(x+6)=0$
$⇔(x+6)(3x-1+x-6)=0$
$⇔(x+6)(4x-7)=0$
⇔\(\left[ \begin{array}{l}x+6=0\\4x-7=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-6\\x=7/4\end{array} \right.\)
$Vậy$ $x=-6$ $hoặc$ $x=7/4$
