a) $\frac{x+1}{x-2}$ -$\frac{5}{x+2}$ =$\frac{12}{x²-4}$ +1(ĐK x khác +-2)
⇔$\frac{(x+1)(x+2)}{(x-2)(x+2)}$ -$\frac{5(x-2)}{(x-2)(x+2)}$ =$\frac{12+x²-4}{(x-2)(x+2)}$
⇔(x+1)(x+2)-5(x-2)=12+x²-4
⇔x²+3x+2-5x+10=12+x²-4
⇔-2x=-4
⇔x=2(loại)
Vậy pt vô nghiệm
b) |2x+6| - x = 3
⇔ |2x+6| = x+3
⇔\(\left[ \begin{array}{l}2x+6=x+3\\2x+6=-x-3\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=-3\end{array} \right.\)