Question

(x+3y/x^2-3xy^2 + 3/x^2+3xy)

Answer 1

$frac{x+3y}{x^2y-3xy^2}+frac{3}{x^2+3xy}=frac{(x+3y)^2}{xy(x-3y)(x+3y)}+frac{3y(x-3y)}{xy(x+3y)(x-3y)}=frac{x^2+9xy}{xy(x-3y)(x+3y)}=\frac{x+9y}{y(x-3y)(x+3y)}$

Answer 2

Благодарю