ГДЗ по алгебре 7 класс Макарычев Задание 450

\[\boxed{\text{450\ (450).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{2^{5} \cdot \left( 2^{3} \right)^{4}}{2^{13}} = \frac{2^{5} \cdot 2^{12}}{2^{13}} =\]

\[= 2^{5 + 12 - 13} = 2^{4} = 16\]

\[\textbf{б)}\ \frac{\left( 5^{8} \right)^{2} \cdot 5^{7}}{5^{22}} = \frac{5^{16} \cdot 5^{7}}{5^{22}} =\]

\[= 5^{16 + 7 - 22} = 5\]

\[\textbf{в)}\ \frac{\left( 2^{5} \right)^{2}}{2^{6} \cdot 4} = \frac{2^{10}}{2^{6} \cdot 2^{2}} = 2^{10 - 6 - 2} =\]

\[= 2^{2} = 4\]

\[\textbf{г)}\ \frac{3^{7} \cdot 27}{\left( 3^{4} \right)^{3}} = \frac{3^{7} \cdot 3^{3}}{3^{12}} = 3^{7 + 3 - 12} =\]

\[= 3^{- 2} = \frac{1}{9}\]

\[\textbf{д)}\ \frac{\left( 5^{2} \right)^{4} \cdot 25}{5^{9}} = \frac{5^{8} \cdot 5^{2}}{5^{9}} =\]

\[= 5^{8 + 2 - 9} = 5\]

\[\textbf{е)}\ \frac{\left( 7^{3} \right)^{3} \cdot 7^{2}}{\left( 7^{5} \right)^{2}} = \frac{7^{9} \cdot 7^{2}}{7^{10}} =\]

\[= 7^{9 + 2 - 10} = 7\]

\[\textbf{ж)}\ \frac{3^{11} \cdot 27}{\left( 3^{4} \right)^{3} \cdot 9} = \frac{3^{11} \cdot 3^{3}}{3^{12} \cdot 3^{2}} = \frac{3^{14}}{3^{14}} =\]

\[= 3^{14 - 14} = 3{^\circ} = 1\]

\[\textbf{з)}\ \frac{\left( 11^{2} \right)^{3}}{11^{2} \cdot 11^{3}} = \frac{11^{6}}{11^{5}} = 11^{6 - 5} = 11\]