Упр.286 ГДЗ Муравин 7 класс (Алгебра)

1) (-2/3)^7 · (-3/2)^8; 4) (5/4)^11 · (0,8)^9;
2) (-3/5)^12 : (-3/5)^9; 5) (-7/9)^4 : (7/9)^3;
3) (1 1/4)^11 : (1 1/4)^9; 6) (2/3)^8 : 32/(27 · 81).

1. $$\left(-\frac{2}{3}\right)^7 \cdot \left(-\frac{3}{2}\right)^8 = -\frac{2^7}{3^7}\cdot \frac{3^8}{2^8} = -\frac{3}{2} = -1{,}5.$$
2. $$\left(-\frac{3}{5}\right)^{12} : \left(-\frac{3}{5}\right)^9 = \left(-\frac{3}{5}\right)^{12-9} = \left(-\frac{3}{5}\right)^3 = -\frac{27}{125}.$$
3. $$\left(1\frac{1}{4}\right)^{11} : \left(1\frac{1}{4}\right)^9 = \left(\frac{5}{4}\right)^{11-9} = \left(\frac{5}{4}\right)^2 = \frac{25}{16} = 1\frac{9}{16}.$$
4. $$\left(\frac{5}{4}\right)^{11}\cdot (0{,}8)^9 = \left(\frac{5}{4}\right)^{11}\cdot \left(\frac{4}{5}\right)^9 = \frac{5^{11}}{4^{11}}\cdot \frac{4^9}{5^9} = \frac{5^2}{4^2} = \frac{25}{16} = 1\frac{9}{16}.$$
5. $$\left(-\frac{7}{9}\right)^4 : \left(\frac{7}{9}\right)^3 = \left(\frac{7}{9}\right)^4 : \left(\frac{7}{9}\right)^3 = \left(\frac{7}{9}\right)^{4-3} = \frac{7}{9}.$$
6. $$\left(\frac{2}{3}\right)^8 : \frac{32}{27\cdot 81} = \frac{2^8}{3^8} : \frac{2^5}{3^7} = \frac{2^8}{3^8}\cdot \frac{3^7}{2^5} = \frac{2^3}{3} = \frac{8}{3} = 2\frac{2}{3}.$$

Ответ

1) $$-1{,}5$$; 2) $$-\frac{27}{125}$$; 3) $$1\frac{9}{16}$$; 4) $$1\frac{9}{16}$$; 5) $$\frac{7}{9}$$; 6) $$2\frac{2}{3}$$.