ВПР 1 №82 Алгебра = ВПТ 5 №9 Математика
Обчисліть
$\frac{{0,6}^{-4}\cdot\left(1\frac{2}{3}\right)^{-6}}{{0,36}^{-5}\cdot\left(2\frac{7}{9}\right)^{-6}}.$
Розв'язок:
$\frac{{0,6}^{-4}\cdot\left(1\frac{2}{3}\right)^{-6}}{{0,36}^{-5}\cdot\left(2\frac{7}{9}\right)^{-6}}= \frac{\left(\frac{3}{5}\right)^{-4}\cdot\left(\frac{5}{3}\right)^{-6}}{\left(\frac{9}{25}\right)^{-5}\cdot\left(\frac{25}{9}\right)^{-6}}=$
$= \frac{\left(\frac{5}{3}\right)^4\cdot\left(\frac{3}{5}\right)^6}{\left(\frac{25}{9}\right)^5\cdot\left(\frac{9}{25}\right)^6}=$
$= \frac{\left(\frac{5}{3}\right)^4\cdot\left(\frac{3}{5}\right)^4\cdot\left(\frac{3}{5}\right)^2}{\left(\frac{25}{9}\right)^5\cdot\left(\frac{9}{25}\right)^5\cdot\frac{9}{25}}=$
$= \frac{\left(\frac{5}{3}\cdot\frac{9}{5}\right)^4\cdot\frac{9}{25}}{\left(\frac{25}{9}\cdot\frac{9}{25}\right)^9\cdot\frac{9}{25}}=\frac{1\cdot\frac{9}{25}}{1\cdot\frac{9}{25}}=1\mathrm{.}$