a, (4x-1).(x-3)-(x-3).(5x+2)=0 b, (x+6).(3x-1)+x^2-36=0

a) `(4x - 1).(x - 3) - (x - 3).(5x + 2) = 0`

`⇔ (x - 3)(4x - 1 -5x - 2) = 0`

`⇔ (x - 3)(-x - 3) = 0`

`⇔` \(\left[ \begin{array}{l}x - 3=0\\-x - 3=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x =3\\-1 . x =3\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x =3\\x =-3\end{array} \right.\)  

Vậy `S = {3 ; -3}`

b) `(x + 6).(3x - 1) + x^2 - 36 = 0`

`⇔ (x + 6).(3x - 1) + (x - 6).(x + 6) = 0`

`⇔ (x + 6)(3x - 1 + x - 6) = 0`

`⇔ (x + 6)(4x - 7) = 0`

`⇔`  \(\left[ \begin{array}{l}x + 6 =0\\4x - 7 =0\end{array} \right.\)  

`⇔`  \(\left[ \begin{array}{l}x  =-6\\x  = \dfrac{7}{4}\end{array} \right.\)  

Vậy `S = { - 6 ; 7/4}`