a) l5xl = x+2

⇔ \(\left[ \begin{array}{l}x+2=5x\\x+2=-5x\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x-5x=-2\\x+5x=-2\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}-4x=-2\\6x=-2\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=\frac{-1}{3}\end{array} \right.\) 

  Vậy S = {$\frac{1}{2}$; $\frac{-1}{3}$}

b) l-3xl = 2x-3

⇔ 2x-3 = 3x

⇔ 2x-3x = 3

⇔ -x = 3

⇔ x = -3

  Vậy S = {-3}

c) lx+4l = x+3

⇔ \(\left[ \begin{array}{l}x+3=x+4\\x+3=-(x+4)\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x+3=x+4\\x+3=-x-4\end{array} \right.\)  

⇔ \(\left[ \begin{array}{l}x-x=4-3\\x+x=-4-3\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}0x=1 (vô lí)\\2x=-7\end{array} \right.\) 

⇔ x = $\frac{-7}{2}$ 

  Vậy S = {$\frac{-7}{2}$}

d) l3-xl = 2x

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

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

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

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

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

  Vậy S = {1; -3}