Упр.8.35 ГДЗ Мордковича 11 класс профильный уровень (Алгебра)

Задача

Рассмотрим вариант решения задания из учебника Мордкович, Семенов 11 класс, Мнемозина: Упростите выражение: 8.35 а)(a3/2-b3/2)/(a1/2+b1/2)*(a-b)/(a+a1/2b1/2+b)+ 2a1/2b1/2; б)(q1/2/(p-p1/2q1/2)+p1/2/(q-p1/2q1/2))*(pq1/2+p1/2q)/(p-q).

Подробный ответ

а)

$$ \frac{a^{3/2}-b^{3/2}}{a^{1/2}+b^{1/2}}\cdot \frac{a-b}{a+a^{1/2}b^{1/2}+b}+2a^{1/2}b^{1/2} $$

Представим разности степеней через корни:

$$ a^{3/2}-b^{3/2}=(a^{1/2}-b^{1/2})(a+b^{1/2}a^{1/2}+b^{1/2}), $$
$$ a-b=(a^{1/2}-b^{1/2})(a^{1/2}+b^{1/2}). $$

Тогда

$$ \frac{a^{3/2}-b^{3/2}}{a^{1/2}+b^{1/2}}\cdot \frac{a-b}{a+a^{1/2}b^{1/2}+b} = \frac{(a^{1/2}-b^{1/2})(a+a^{1/2}b^{1/2}+b)}{a^{1/2}+b^{1/2}} \cdot \frac{(a^{1/2}-b^{1/2})(a^{1/2}+b^{1/2})}{a+a^{1/2}b^{1/2}+b} $$

$$ =(a^{1/2}-b^{1/2})^2. $$

Следовательно,

$$ (a^{1/2}-b^{1/2})^2+2a^{1/2}b^{1/2} = a-2a^{1/2}b^{1/2}+b+2a^{1/2}b^{1/2} = a+b. $$

б)

$$ \left(\frac{q^{1/2}}{p-p^{1/2}q^{1/2}}+\frac{p^{1/2}}{q-p^{1/2}q^{1/2}}\right)\cdot \frac{pq^{1/2}+p^{1/2}q}{p-q} $$

Преобразуем знаменатели:

$$ p-p^{1/2}q^{1/2}=p^{1/2}(p^{1/2}-q^{1/2}), \qquad q-p^{1/2}q^{1/2}=q^{1/2}(q^{1/2}-p^{1/2}). $$

Тогда

$$ \frac{q^{1/2}}{p^{1/2}(p^{1/2}-q^{1/2})} +\frac{p^{1/2}}{q^{1/2}(q^{1/2}-p^{1/2})} = \frac{q-p}{p^{1/2}q^{1/2}(p^{1/2}-q^{1/2})}. $$

Кроме того,

$$ pq^{1/2}+p^{1/2}q=p^{1/2}q^{1/2}(p^{1/2}+q^{1/2}), $$
$$ p-q=(p^{1/2}-q^{1/2})(p^{1/2}+q^{1/2}). $$

Тогда всё выражение равно

$$ \frac{q-p}{p^{1/2}q^{1/2}(p^{1/2}-q^{1/2})}\cdot \frac{p^{1/2}q^{1/2}(p^{1/2}+q^{1/2})}{(p^{1/2}-q^{1/2})(p^{1/2}+q^{1/2})} = \frac{q-p}{(p^{1/2}-q^{1/2})^2}. $$

Так как $$q-p=-(p-q)$$ и $$p-q=(p^{1/2}-q^{1/2})(p^{1/2}+q^{1/2})$$, получаем

$$ \frac{q-p}{(p^{1/2}-q^{1/2})^2} = \frac{q^{1/2}+p^{1/2}}{q^{1/2}-p^{1/2}}. $$