Вычислите: $\frac{1}{24} + \frac{2}{3}$

1. a) $$\frac{1}{24} + \frac{2}{3} = \frac{1}{24} + \frac{2 \cdot 8}{3 \cdot 8} = \frac{1}{24} + \frac{16}{24} = \frac{1+16}{24} = \frac{17}{24}$$ b) $$\left(\frac{1}{3}\right)^2 + (-3)^3 = \frac{1^2}{3^2} + (-3 \cdot -3 \cdot -3) = \frac{1}{9} - 27 = \frac{1}{9} - \frac{27 \cdot 9}{9} = \frac{1}{9} - \frac{243}{9} = \frac{1-243}{9} = -\frac{242}{9} = -26\frac{8}{9}$$ 2. а) $$\frac{8}{9} - 8^2 \cdot \frac{1}{72} = \frac{8}{9} - 64 \cdot \frac{1}{72} = \frac{8}{9} - \frac{64}{72} = \frac{8 \cdot 8}{9 \cdot 8} - \frac{64}{72} = \frac{64}{72} - \frac{64}{72} = 0$$ b) $$\left(-2\frac{1}{4}\right)^2 + (-2)^3 = \left(-\frac{9}{4}\right)^2 + (-8) = \frac{81}{16} - 8 = \frac{81}{16} - \frac{8 \cdot 16}{16} = \frac{81}{16} - \frac{128}{16} = \frac{81-128}{16} = -\frac{47}{16} = -2\frac{15}{16}$$ **Задание 2: Найдите значение выражения** a) $$\text{0,5}x^5 \text{ при } x = -2$$ $$0,5 \cdot (-2)^5 = 0,5 \cdot (-32) = -16$$ b) $$1 - \frac{1}{27}x^3 \text{ при } x = -3$$ $$1 - \frac{1}{27} \cdot (-3)^3 = 1 - \frac{1}{27} \cdot (-27) = 1 - (-1) = 1 + 1 = 2$$ **Задание 3: Упростите выражения** a) $$\frac{(c^3)^5 \cdot c^2}{c^{17}} = \frac{c^{3 \cdot 5} \cdot c^2}{c^{17}} = \frac{c^{15} \cdot c^2}{c^{17}} = \frac{c^{15+2}}{c^{17}} = \frac{c^{17}}{c^{17}} = c^{17-17} = c^0 = 1$$ b) $$\frac{(a^2 \cdot a^3)^3}{a^7} = \frac{(a^{2+3})^3}{a^7} = \frac{(a^5)^3}{a^7} = \frac{a^{5 \cdot 3}}{a^7} = \frac{a^{15}}{a^7} = a^{15-7} = a^8$$ v) $$(-6abc)^3 = (-6)^3 a^3 b^3 c^3 = -216 a^3 b^3 c^3$$ г) $$\frac{(c^5)^4 \cdot c^7}{c^{27}} = \frac{c^{5 \cdot 4} \cdot c^7}{c^{27}} = \frac{c^{20} \cdot c^7}{c^{27}} = \frac{c^{20+7}}{c^{27}} = \frac{c^{27}}{c^{27}} = c^{27-27} = c^0 = 1$$ д) $$\frac{(a^3 \cdot a^2)^4}{a^{19}} = \frac{(a^{3+2})^4}{a^{19}} = \frac{(a^5)^4}{a^{19}} = \frac{a^{5 \cdot 4}}{a^{19}} = \frac{a^{20}}{a^{19}} = a^{20-19} = a^1 = a$$ e) $$(-7xyz)^3 = (-7)^3 x^3 y^3 z^3 = -343 x^3 y^3 z^3$$ **Задание 4: Вычислите, используя свойства степени** a) $$1,3^4 \cdot \left(\frac{10}{13}\right)^4 = \left(1,3 \cdot \frac{10}{13}\right)^4 = \left(\frac{13}{10} \cdot \frac{10}{13}\right)^4 = (1)^4 = 1$$ b) $$2,1^7 \cdot \left(\frac{10}{21}\right)^7 = \left(2,1 \cdot \frac{10}{21}\right)^7 = \left(\frac{21}{10} \cdot \frac{10}{21}\right)^7 = (1)^7 = 1$$ v) $$\frac{100^3 \cdot 10^7}{2^{13} \cdot 5^{13}} = \frac{(10^2)^3 \cdot 10^7}{(2 \cdot 5)^{13}} = \frac{10^{2 \cdot 3} \cdot 10^7}{10^{13}} = \frac{10^6 \cdot 10^7}{10^{13}} = \frac{10^{6+7}}{10^{13}} = \frac{10^{13}}{10^{13}} = 10^{13-13} = 10^0 = 1$$ г) $$\frac{36^6 \cdot 6^4}{2^{10} \cdot 3^{16}} = \frac{(6^2)^6 \cdot 6^4}{2^{10} \cdot 3^{16}} = \frac{6^{12} \cdot 6^4}{2^{10} \cdot 3^{16}} = \frac{6^{12+4}}{2^{10} \cdot 3^{16}} = \frac{6^{16}}{2^{10} \cdot 3^{16}} = \frac{(2 \cdot 3)^{16}}{2^{10} \cdot 3^{16}} = \frac{2^{16} \cdot 3^{16}}{2^{10} \cdot 3^{16}} = 2^{16-10} \cdot 3^{16-16} = 2^6 \cdot 3^0 = 64 \cdot 1 = 64$$