ДОКАЖИТЕ ТОЖДЕСТВО: 4x^2. 12x^3 2x^2. 1 _____ : ________ * ______ = ____ 2x-y 4x^2-y^2....

Решите задачу:

$\frac{4x^2}{2x-y}: \frac{12x^3}{4x^2-y^2}* \frac{2x^2}{6x^2+3xy}= \frac{1}{9} \\\frac{4x^2}{(2x-y)}* \frac{(2x-y)(2x+y)}{12x^3}* \frac{2x^2}{6x^2+3xy}= \frac{1}{9} \\\frac{2x+y}{3x}* \frac{2x^2}{6x^2+3xy}= \frac{1}{9} \\\frac{2x+y}{3x}* \frac{2x^2}{3x(2x+y)}= \frac{1}{9} \\ \frac{2x^2}{9x^2}= \frac{1}{9} \\ \frac{2}{9} \neq \frac{1}{9}$