Упр.8.35 ГДЗ Мордковича 11 класс профильный уровень (Алгебра)
- 8.35. Упростите выражение:
а) $$\frac{a^{3/2}-b^{3/2}}{a^{1/2}+b^{1/2}}\cdot\frac{a-b}{a+a^{1/2}b^{1/2}+b}+2a^{1/2}b^{1/2}$$;
б) $$\left(\frac{q^{1/2}}{p-p^{1/2}q^{1/2}}+\frac{p^{1/2}}{q-p^{1/2}q^{1/2}}\right)\cdot\frac{pq^{1/2}+p^{1/2}q}{p-q}$$.
а) Преобразуем выражение, используя формулы разности квадратов и разности кубов:
$$\frac{a^{\frac32}-b^{\frac32}}{a^{\frac12}+b^{\frac12}}\cdot \frac{a-b}{a+a^{\frac12}b^{\frac12}+b} +2a^{\frac12}b^{\frac12}$$
$$= \frac{\left(a^{\frac12}-b^{\frac12}\right)\left(a+b^{\frac12}a^{\frac12}+b\right)}{a^{\frac12}+b^{\frac12}} \cdot \frac{\left(a^{\frac12}-b^{\frac12}\right)\left(a^{\frac12}+b^{\frac12}\right)}{a+a^{\frac12}b^{\frac12}+b} +2a^{\frac12}b^{\frac12}$$
$$=\left(a^{\frac12}-b^{\frac12}\right)^2+2a^{\frac12}b^{\frac12}$$
$$=a-2a^{\frac12}b^{\frac12}+b+2a^{\frac12}b^{\frac12}=a+b.$$
Ответ: $$a+b$$.
б) Упростим выражение:
$$\left(\frac{q^{\frac12}}{p-p^{\frac12}q^{\frac12}}+\frac{p^{\frac12}}{q-p^{\frac12}q^{\frac12}}\right)\cdot \frac{pq^{\frac12}+p^{\frac12}q}{p-q}$$
$$= \left(\frac{q^{\frac12}}{p^{\frac12}\left(p^{\frac12}-q^{\frac12}\right)}+\frac{p^{\frac12}}{q^{\frac12}\left(q^{\frac12}-p^{\frac12}\right)}\right)\cdot \frac{p^{\frac12}q^{\frac12}\left(p^{\frac12}+q^{\frac12}\right)}{\left(p^{\frac12}-q^{\frac12}\right)\left(p^{\frac12}+q^{\frac12}\right)}$$
$$= \left(\frac{q^{\frac12}}{p^{\frac12}\left(p^{\frac12}-q^{\frac12}\right)}-\frac{p^{\frac12}}{q^{\frac12}\left(p^{\frac12}-q^{\frac12}\right)}\right)\cdot \frac{p^{\frac12}q^{\frac12}}{p^{\frac12}-q^{\frac12}}$$
$$= \frac{q-p}{\left(p^{\frac12}-q^{\frac12}\right)^2} = \frac{\left(q^{\frac12}-p^{\frac12}\right)\left(q^{\frac12}+p^{\frac12}\right)}{\left(p^{\frac12}-q^{\frac12}\right)^2} = \frac{q^{\frac12}+p^{\frac12}}{q^{\frac12}-p^{\frac12}}.$$
Ответ: $$\frac{\sqrt{q}+\sqrt{p}}{\sqrt{q}-\sqrt{p}}$$.
