Упр.26.13 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

Задача

1) (a^(2/3)-2a^(1/3)b^(1/3)+b^(2/3))/(a^(4/3)-a^(2/3)b^(2/3)):(a^(1/3)b^(1/3)-b^(2/3))/(ab^(1/3)+a^(2/3)b^(2/3));
2) (a-b)/(a^(3/4)+a^(1/2)b^(1/4))·(a-b)/((a^(1/2)+b^(1/2))(a^(1/2)-a^(1/4)b^(1/4))).

Подробный ответ

1) Преобразуем выражение:

$$ \frac{a^{2/3}-2a^{1/3}b^{1/3}+b^{2/3}}{a^{4/3}-a^{2/3}b^{2/3}}:\frac{a^{1/3}b^{1/3}-b^{2/3}}{ab^{1/3}+a^{2/3}b^{2/3}} $$

$$ a^{2/3}-2a^{1/3}b^{1/3}+b^{2/3}=(a^{1/3}-b^{1/3})^2, $$
$$ a^{4/3}-a^{2/3}b^{2/3}=a^{2/3}(a^{2/3}-b^{2/3})=a^{2/3}(a^{1/3}-b^{1/3})(a^{1/3}+b^{1/3}), $$
$$ a^{1/3}b^{1/3}-b^{2/3}=b^{1/3}(a^{1/3}-b^{1/3}), $$
$$ ab^{1/3}+a^{2/3}b^{2/3}=a^{2/3}b^{1/3}(a^{1/3}+b^{1/3}). $$

Тогда

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

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

2) Упростим выражение:

$$ \frac{a-b}{a^{3/4}+a^{1/2}b^{1/4}}\cdot \frac{a-b}{(a^{1/2}+b^{1/2})(a^{1/2}-a^{1/4}b^{1/4})} $$

$$ a-b=(a^{1/2}-b^{1/2})(a^{1/2}+b^{1/2}), $$
$$ a^{3/4}+a^{1/2}b^{1/4}=a^{1/2}(a^{1/4}+b^{1/4}), $$
$$ a^{1/2}-a^{1/4}b^{1/4}=a^{1/4}(a^{1/4}-b^{1/4}). $$

Тогда

$$ \frac{(a^{1/2}-b^{1/2})^2(a^{1/2}+b^{1/2})^2} {a^{1/2}(a^{1/4}+b^{1/4})(a^{1/2}+b^{1/2})a^{1/4}(a^{1/4}-b^{1/4})} $$

$$ = \frac{(a^{1/2}-b^{1/2})^2(a^{1/2}+b^{1/2})} {a^{3/4}(a^{1/4}+b^{1/4})(a^{1/4}-b^{1/4})}. $$

Используем формулу разности квадратов:

$$ (a^{1/4}+b^{1/4})(a^{1/4}-b^{1/4})=a^{1/2}-b^{1/2}. $$

Получаем

$$ \frac{(a^{1/2}-b^{1/2})(a^{1/2}+b^{1/2})}{a^{3/4}} = \frac{a-b}{a^{3/4}} = a^{1/4}-ba^{-3/4}. $$