$1) x² -1 = 0$
$⇔ x² = 1$
$⇔ x = ±1$
Vậy S = {$±1$}
$2) \frac{x +1}{x -2} = \frac{1}{x² -4}$ $(ĐKXĐ:$ $x \neq ±2)$
$⇔ \frac{(x +1).(x +2)}{(x -2).(x +20} = \frac{1}{(x -2).(x +2)}$
$⇒ (x +1).(x +2)=1$
$⇔ x² +3x +2 -1=0$
$⇔ x² +3x +1=0$
$⇔ (x +\frac{3}{2})² - \frac{5}{4} = 0$
$⇔ (x +\frac{3}{2} - \frac{√5}{2}).(x +\frac{3}{2} + \frac{√5}{2}) = 0$
$⇔ \left[ \begin{array}{l}x +\frac{3}{2} - \frac{√5}{2}=0\\x +\frac{3}{2} + \frac{√5}{2}=0\end{array} \right.$ $⇔ \left[ \begin{array}{l}x=\frac{-3 +√5}{2}\\x=\frac{-3-√5}{2}\end{array} \right.$ $(T/m$ $đkxđ)$
$Vậy$ $S =$ {$\frac{-3 +√5}{2}; \frac{-3-√5}{2}$}
$3) \frac{4x+2}{x-11} = \frac{12x+5}{3x+4}$ $(ĐKXĐ:$ $x \neq 11; x \neq \frac{-4}{3})$
$⇔ \frac{(4x+2).(3x +4)}{(x-11).(3x +4)} = \frac{(12x+5).(x -11)}{(3x+4).(x -11)}$
$⇒ (4x+2).(3x +4) = (12x+5).(x -11)$
$⇔ 12x² +22x +8 = 12x² -137x-55$
$⇔ 12x² -12x² +22x +137x = -55 -8$
$⇔ 159x = -63$
$⇔ x = \frac{-21}{53}$ $(T/m$ $đkxđ)$
$Vậy$ $S =$ {$\frac{-21}{53}$}
