Упр.45 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1. Упростить выражение: 1) $$\left(a^6b^7\right)^{\frac{1}{5}}:\left(ab^2\right)^{\frac{1}{5}}$$; 2) $$\left(81x^4y\right)^{\frac{1}{3}}:\left(3xy\right)^{\frac{1}{3}}$$; 3) $$\left(\frac{3x}{y^2}\right)^{\frac{1}{3}}:\left(\frac{y}{9x^2}\right)^{\frac{1}{3}}$$; 4) $$\left(\frac{2b}{a^3}\right)^{\frac{1}{4}}:\left(\frac{a}{8b^3}\right)^{\frac{1}{4}}$$.

1. $$\left(a^6b^7\right)^{\frac15}:\left(ab^2\right)^{\frac15} =\left(\frac{a^6b^7}{ab^2}\right)^{\frac15} =\left(a^5b^5\right)^{\frac15} =ab.$$

2. $$\left(81x^4y\right)^{\frac13}:\left(3xy\right)^{\frac13} =\left(\frac{81x^4y}{3xy}\right)^{\frac13} =\left(27x^3\right)^{\frac13} =3x.$$

3. $$\left(\frac{3x}{y^2}\right)^{\frac13}:\left(\frac{y}{9x^2}\right)^{\frac13} =\left(\frac{3x}{y^2}\cdot\frac{9x^2}{y}\right)^{\frac13} =\left(\frac{27x^3}{y^3}\right)^{\frac13} =\frac{3x}{y}.$$

4. $$\left(\frac{2b}{a^3}\right)^{\frac14}:\left(\frac{a}{8b^3}\right)^{\frac14} =\left(\frac{2b}{a^3}\cdot\frac{8b^3}{a}\right)^{\frac14} =\left(\frac{16b^4}{a^4}\right)^{\frac14} =\frac{2b}{a}.$$

Ответ

1) $$ab$$; 2) $$3x$$; 3) $$\frac{3x}{y}$$; 4) $$\frac{2b}{a}$$.