Упр.93 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

Задача

1) (a^(1/4)-a^(-7/4))/(a^(1/4)-a^(-3/4));
2) (a^(4/3)-a^(-2/3))/(a^(1/3)-a6(-2/3));
3) (b^(5/4)+2b^(1/4)+b^(-3/4))/(b^(3/4)+b^(-1/4));
4) (a^(-4/3)b^(-2)-a^(-2)b^(-4/3))/(a^(-5/3)b^(-2)-b^(-5/3)a^(-2));
5) (v(a^3 b^(-1))-v(a^(-1)b^3))/(v(ab^(-1))-v(a^(-1) b));
6) (a^(3/4)b^(-1/4)-a^(-1/4)b^(3/4))/(a^(1/4)b^(-1/4)+a^(-1/4)b^(1/4)).

Подробный ответ

$$\frac{a^{1/4}-a^{-7/4}}{a^{1/4}-a^{-3/4}}= \frac{a^{1/4}(1-a^{-2})}{a^{1/4}(1-a^{-1})}= \frac{(1+a^{-1})(1-a^{-1})}{1-a^{-1}}=1+\frac{1}{a}.$$
$$\frac{a^{4/3}-a^{-2/3}}{a^{1/3}-a^{-2/3}}= \frac{a^{-2/3}(a^2-1)}{a^{-2/3}(a-1)}= \frac{(a-1)(a+1)}{a-1}=a+1.$$
$$\frac{b^{5/4}+2b^{1/4}+b^{-3/4}}{b^{3/4}+b^{-1/4}}= \frac{b^{1/4}(b+2+b^{-1})}{b^{-1/4}(b+1)}= \frac{b+2+\frac{1}{b}}{\sqrt{b}+\frac{1}{\sqrt{b}}}.$$
Так как
$$b+2+\frac{1}{b}=\left(\sqrt{b}+\frac{1}{\sqrt{b}}\right)^2,$$
то
$$\frac{b+2+\frac{1}{b}}{\sqrt{b}+\frac{1}{\sqrt{b}}}=\sqrt{b}+\frac{1}{\sqrt{b}}.$$
$$\frac{a^{-4/3}b^{-2}-a^{-2}b^{-4/3}}{a^{-5/3}b^{-2}-b^{-5/3}a^{-2}}= \frac{a^{-2}b^{-2}\left(a^{2/3}-b^{2/3}\right)}{a^{-2}b^{-2}\left(a^{1/3}-b^{1/3}\right)}.$$
Используем формулу разности квадратов:
$$a^{2/3}-b^{2/3}=\left(a^{1/3}-b^{1/3}\right)\left(a^{1/3}+b^{1/3}\right).$$
Тогда
$$\frac{a^{2/3}-b^{2/3}}{a^{1/3}-b^{1/3}}=a^{1/3}+b^{1/3}.$$
$$\frac{\sqrt{a^3b^{-1}}-\sqrt{a^{-1}b^3}}{\sqrt{ab^{-1}}-\sqrt{a^{-1}b}}= \frac{\sqrt{a^{-1}b^{-1}}\left(\sqrt{a^4}-\sqrt{b^4}\right)}{\sqrt{a^{-1}b^{-1}}\left(\sqrt{a^2}-\sqrt{b^2}\right)}.$$
После сокращения получаем:
$$\frac{a^2-b^2}{a-b}=\frac{(a-b)(a+b)}{a-b}=a+b.$$
$$\frac{a^{3/4}b^{-1/4}-a^{-1/4}b^{3/4}}{a^{1/4}b^{-1/4}+a^{-1/4}b^{1/4}}= \frac{a^{-1/4}b^{-1/4}(a- b)}{a^{-1/4}b^{-1/4}(\sqrt{a}+\sqrt{b})}.$$
Преобразуем числитель:
$$a^{3/4}b^{-1/4}-a^{-1/4}b^{3/4}=a^{-1/4}b^{-1/4}(a-b),$$
а знаменатель:
$$a^{1/4}b^{-1/4}+a^{-1/4}b^{1/4}=a^{-1/4}b^{-1/4}(\sqrt{a}+\sqrt{b}).$$
Тогда
$$\frac{a-b}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}.$$