Для решения уравнения (3x-1)(2x+4)-(x+1)(6x-5)=16 выполним следующие шаги:
$$ (3x-1)(2x+4) = 3x \cdot 2x + 3x \cdot 4 - 1 \cdot 2x - 1 \cdot 4 = 6x^2 + 12x - 2x - 4 = 6x^2 + 10x - 4 $$
$$ (x+1)(6x-5) = x \cdot 6x + x \cdot (-5) + 1 \cdot 6x + 1 \cdot (-5) = 6x^2 - 5x + 6x - 5 = 6x^2 + x - 5 $$
$$ (6x^2 + 10x - 4) - (6x^2 + x - 5) = 16 $$
$$ 6x^2 + 10x - 4 - 6x^2 - x + 5 = 16 $$
$$ (6x^2 - 6x^2) + (10x - x) + (-4 + 5) = 16 $$
$$ 9x + 1 = 16 $$
$$ 9x = 16 - 1 $$
$$ 9x = 15 $$
$$ x = \frac{15}{9} $$
$$ x = \frac{5}{3} $$
Ответ: $$ x = \frac{5}{3} $$