Question

2(p+q)/p^3-q^3+2/q^2-p^2

Answer 1

Решение и ответ:

$\displaystyle\frac{{2(p+q)}}{{{p^3}-{q^3}}}+\frac{2}{{{q^2}-{p^2}}}=\frac{{2(p+q)}}{{(p-q)({p^2}+pq+{q^2})}}+\frac{2}{{(q-p)(q+p)}}=$

$\displaystyle\frac{{2(p+q)}}{{(p-q)({p^2}+pq+{q^2})}}-\frac{2}{{(p-q)(q+p)}}=\frac{{2(p+q)(p+q)-2({p^2}+pq+{q^2})}}{{(p-q)({p^2}+pq+{q^2})(q+p)}}=$

$\displaystyle\frac{{2((p+q)(p+q)-({p^2}+pq+{q^2}))}}{{(p-q)({p^2}+pq+{q^2})(q+p)}}=\frac{{2(({p^2}+pq+qp+{q^2})-({p^2}+pq+{q^2}))}}{{(p-q)({p^2}+pq+{q^2})(q+p)}}=$

$\displaystyle\frac{{2({p^2}+pq+qp+{q^2}-{p^2}-pq-{q^2})}}{{(p-q)({p^2}+pq+{q^2})(q+p)}}=\frac{{2qp}}{{(p-q)({p^2}+pq+{q^2})(q+p)}}$