$b) (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)$
$⇔ (3x – 1)(x2 + 2) – (3x – 1)(7x – 10) = 0$
$⇔ (3x – 1)(x2 + 2 – 7x + 10) = 0$
$⇔ (3x – 1)(x2 – 7x + 12) = 0$
$⇔ (3x – 1)(x2 – 4x – 3x + 12) = 0$
$⇔ (3x – 1)[(x2 – 4x) – (3x - 12)] = 0$
$⇔ (3x – 1)[x(x – 4) – 3(x – 4)] = 0$
$⇔ (3x – 1)(x – 3)(x – 4) = 0$
$⇔ 3x – 1 = 0$ hoặc $x – 3 = 0$
$+ 3x – 1 = 0 ⇔ 3x = 1 ⇔ x = 1/3.$
$+ x – 3 = 0 ⇔ x = 3.$
$+ x – 4 = 0 ⇔ x = 4.$
Vậy phương trình có tập nghiệm là $S=[$$\frac{1}{3};3;4]$
Xin CTRLHN
