Giải phương trình: $\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}$

$\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}$

⇔ 4(x – 2)(x + 10) – (x + 4)(x + 10) = 3(x – 2)(x + 4)

⇔ (x – 2)(x + 10) – (x + 4)(x + 10) + 3(x – 2)(x + 10) – 3(x – 2)(x + 4) = 0

⇔ (x + 10) (x – 2 – x – 4) + 3(x – 2)(x + 10 – x – 4) = 0

⇔ (x + 10).(–6) + 3(x – 2).6 = 0

⇔ 6.(–x – 10 + 3x – 6) = 0

⇔ 6.(2x – 16) = 0

⇔ 2x – 16 = 0

⇔ 2x = 16

⇔ x = 8

Vậy x = 8.