a. $(x^{2} +1)(x^{2} - 4x +4) = 0$
⇔ $(x^{2} +1)(x-2)^{2}=0$
⇔\(\left[ \begin{array}{l}x^{2}+1=0 ( vô lí)\\x-2=0\end{array} \right.\)
⇔ $x-2=0$
⇔$x=2$
S = { 2 }
b. $(3x-2)(\frac{2(x+3)}{7} - \frac{4x-3}{5}=0$
⇔ $(3x-2)(\frac{51-18x}{35})=0$
⇔ $\(\left[ \begin{array}{l}3x-2=0\\51-18x=0\end{array} \right.\)$
⇔ $\(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=\dfrac{17}{6}\end{array} \right.\) $
$S =$ { $\dfrac{2}{3}$ ; $\dfrac{17}{6}$ )
c. $(3,3-11x)(\frac{7x+2}{5} + \frac{2(1-3x)}{3})=0$
⇔ $(3,3-11x)(\frac{21x+6+10-30}{15})=0$
⇔$(3,3-11x)(\frac{16-9x}{15}=0$
⇔$ \(\left[ \begin{array}{l}3,3-11x=0\\16-9x=0\end{array} \right.\) $
⇔$ \(\left[ \begin{array}{l}x=\dfrac{3}{10}\\x=\dfrac{16}{9}\end{array} \right.\) $
$S$ = { $\dfrac{3}{10}$ ; $\dfrac{16}{9}$ }
