Đáp án:
$a, (x+2)(3x-7)=0$
\(⇔\left[ \begin{array}{l} x + 2=0\\3x - 7=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=-2\\x=\frac{7}{3}\end{array} \right.\)
$b,(x-2)(2x-3)(7-2x)=0$
$1) x - 2 = 0 ⇔ x = 2$
$2) 2x - 3 = 0 ⇔ x = \frac{3}{2}$
$ 3) 7 - 2x = 0 ⇔ x = \frac{7}{2}$
$c,x³-5x²+6x=0$
$⇔ x³ - 2x² - 3x² + 6x = 0$
$⇔ x²(x - 2) - 3x(x - 2) = 0$
$⇔ (x² - 3x)(x - 2) = 0$
$⇔ x(x - 3)(x - 2) = 0$
$1) x = 0$
$2) x - 3 = 0 ⇔ x = 3$
$3) x - 2 = 0 ⇔ x = 2$
