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;. Created by jakesmit3. algebra-ru

6x-6y;...

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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;

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instant196 $\frac{p^2-4q^2}{(p+2q)^2} = \frac{(p-2q)(p+2q)}{(p+2q)^2} =\frac{p-2q}{p+2q}$
$\frac{ p^3-8q^3}{4q^2+2pq+p^2} = \frac{(p-2q)(4q^2+2pq+p^2)}{4q^2+2pq+p^2} = p-2q$
$\frac{3x^2-3y^2}{x^2+xy} * \frac{x+y}{6x-6y} = \frac{3(x+y)^{2}(x-y) }{6x(x+y)(x-y)} = \frac{x+y}{2x}$

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