Question

Помогите решить! 4y^3/x-y^2 + 3xy+y^3/y^2-x

Answer 1

$\frac{4y^3}{x-y^2}+ \frac{3xy+y^3}{y^2-x}= \frac{4y^3}{x-y^2}- \frac{3xy+y^3}{x-y^2}= \frac{4y^3-3xy-y^3}{x-y^2}= \frac{3y^3-3xy}{x-y^2}=$
$\frac{3y(y^2-x)}{x-y^2}= -\frac{3y(x-y^2)}{x-y^2}=-3y$