7) $$\frac{x-12y}{x^2-16y^2} - \frac{4y}{4xy-x^2}$$

7) $$\frac{x-12y}{x^2-16y^2} - \frac{4y}{4xy-x^2}$$ Разложим знаменатели: $$x^2-16y^2 = (x-4y)(x+4y)$$ $$4xy-x^2 = x(4y-x) = -x(x-4y)$$ Тогда: $$\frac{x-12y}{(x-4y)(x+4y)} - \frac{4y}{-x(x-4y)} = \frac{x-12y}{(x-4y)(x+4y)} + \frac{4y}{x(x-4y)} = \frac{x(x-12y)+4y(x+4y)}{x(x-4y)(x+4y)} =$$ $$=\frac{x^2-12xy+4xy+16y^2}{x(x-4y)(x+4y)} = \frac{x^2-8xy+16y^2}{x(x-4y)(x+4y)} = \frac{(x-4y)^2}{x(x-4y)(x+4y)} = \frac{x-4y}{x(x+4y)}$$ Ответ: $$\frac{x-4y}{x(x+4y)}$$