`a) (3x−4)(x−11) = (x+1)(2x+3)`

`⇔ 3x^2-37x+44=2x^2+5x+3`

`⇔3x^2-37x-2x^2-5x+44-3=0`

`⇔x^2-42x+41=0`

`⇔ (x^2-1)-(42x-42)=0`

`⇔(x-1)(x+1)-42(x-1)=0`

`⇔(x-1)(x+1-42)=0`

`⇔(x-1)(x-41)=0`

`⇔` \(\left[ \begin{array}{l}x-1=0\\x-41=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=1\\x=41\end{array} \right.\) 

                                 Vậy `S={1; 41}`

`b) 4/(x-3) - (x+2)/(x+3) + 6/(x^2-9) = -2`    `(đk: xne+-3)`

`⇔4/(x-3) - (x+2)/(x+3) + 6/((x-3)(x+3)) = -2`

`⇒ 4(x+3)-(x+2)(x-3)+6=-2(x-3)(x+3)`

`⇔4x+12-x^2+x+6+6=-2(x^2-9)`

`⇔-x^2+5x+24=-2x^2+18`

`⇔-x^2+5x+24+2x^2-18=0`

`⇔x^2+5x+6=0`

`⇔ (x^2-4)+(5x+10)=0`

`⇔(x-2)(x+2)+5(x+2)=0`

`⇔(x+2)(x-2+5)=0`

`⇔(x+2)(x+3)=0`

`⇔` \(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=-2\\x=-3\end{array} \right.\) 

   Loại `x=-3` vì `ktm đk`

                         Vậy `S={-2}`