Question

Помогите решить:
p^2-4q^2//(p+2q)^2 : p^3-8q^3//4q^2+2pq+p^2;
3x^2-3y^2//x^2+xy * x+y//6x-6y;

Answer 1

$frac{p^2-4q^2}{(p+2q)^2} = frac{(p-2q)(p+2q)}{(p+2q)^2} =frac{p-2q}{p+2q}$
$frac{ p^3-8q^3}{4q^2+2pq+p^2} = frac{(p-2q)(4q^2+2pq+p^2)}{4q^2+2pq+p^2} = p-2q$
$frac{3x^2-3y^2}{x^2+xy} * frac{x+y}{6x-6y} = frac{3(x+y)^{2}(x-y) }{6x(x+y)(x-y)} = frac{x+y}{2x}$