Giải các pt sau: a/3x^2 + 12x – 66 = 0 b/9x^2 -30x +225 = 0 c/3x^2 – 7x + 1 = 0

$a) 3x² + 12x – 66 = 0$

$⇔ 3.(x² +4x -22) = 0$

$⇔ 3.[(x +2)² -26] = 0$

$⇔ 3.(x +2 -√26).(x +2 +√26) = 0$

$⇔ \left[ \begin{array}{l}x+2 -√26=0\\x+2 +√26=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-2 +√26\\x=-2 -√26\end{array} \right.$

$Vậy$ $S =$ {$-2 +√26; -2 -√26$}

$b) 9x² -30x +225 = 0$

$⇔ 3.(3x² -10x +75) = 0$

$⇔ 3.[(√3x -\frac{5√3}{3})² + \frac{200}{3}] = 0$ $(Vô$ $lí)$

$Vì$ $(√3x -\frac{5√3}{3})² + \frac{200}{3} > 0$ $(Vs$ $∀$ $x)$

$Vậy$ $S = ∅$

$c) 3x² – 7x + 1 = 0$

$⇔ (√3x -\frac{7√3}{6})² – \frac{37}{12} = 0$

$⇔ (√3x -\frac{7√3}{6} – \sqrt{\frac{37}{12} }).(√3x -\frac{7√3}{6} + \sqrt{\frac{37}{12} }) = 0$

$⇔ \left[ \begin{array}{l}√3x -\frac{7√3}{6} – \sqrt{\frac{37}{12} }=0\\√3x -\frac{7√3}{6} + \sqrt{\frac{37}{12} }=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\frac{7 +√37}{6}\\x=\frac{7 -√37}{6}\end{array} \right.$

$Vậy$ $S =$ {$\frac{7 +√37}{6}; \frac{7 -√37}{6}$}