25 Б. помогите4x^2-6xy+9y^2 9y^2-4x^2-------------------- * ------------------ = ( 2x-3y...

Решите задачу:

$\frac{4x^2-6xy+9y^2}{2x-3y}* \frac{9y^2-4x^2}{8x^3+27y^3} = \frac{(2x)^2-2x*3y+(3y)^2}{2x-3y}* \frac{(3y)^2-(2x)^2}{(2x)^3+(3y)^3} = \\ \frac{(2x)^2-2x*3y+(3y)^2}{2x-3y}* \frac{(3y-2x)(3y+2x)}{(2x+3y)((2x)^2-2x*3y+(3y)^2)} = \\ \frac{1}{2x-3y}* \frac{(3y-2x)(3y+2x)}{(2x+3y)} = \frac{1}{2x-3y}* \frac{(3y-2x)}{1} = \\ \frac{1}{2x-3y}* -(2x-3y) = -1$