Решебник по алгебре 7 класс Макарычев ФГОС Задание 535

\[\boxed{\text{535.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

Пояснение.

Решение.

\[\textbf{а)}\ ( - 0,03)^{8} = {0,03}^{8} > 0\]

\[\textbf{б)}\ ( - 1,25)^{7} < 0\]

\[\textbf{в)}\ ( - 1,75)^{3} < ( - 0,29)^{2}\]

\[\textbf{г)}\ {0,98}^{6} < {1,02}^{6};\ \ \ \ \]

\[\ так\ как\ 0,98 < 1,02.\]

\[\boxed{\text{535\ (535).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 13^{100}\ :13^{98} = 13^{100 - 98} =\]

\[= 13^{2} = 169\]

\[\textbf{б)}\ \frac{3^{8} \cdot 2^{7}}{3^{6} \cdot 2^{5}} = 3^{8 - 6} \cdot 2^{7 - 5} =\]

\[= 3^{2} \cdot 2^{2} = 9 \cdot 4 = 36\]

\[\textbf{в)}\ 2^{14}\ :8^{4} = 12^{14}\ :\left( 2^{3} \right)^{4} =\]

\[= 2^{14}\ :2^{12} = 2^{14 - 12} = 2^{2} = 4\]

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

\[= \frac{3^{10 - 9}}{5^{10 - 9}} = \frac{3}{5}\]

\[\textbf{д)}\ 5^{10}\ :25^{4} = 5^{10}\ :\left( 5^{2} \right)^{4} =\]

\[= 5^{10}\ :5^{8} = 5^{10 - 8} = 5^{2} = 25\]

\[\textbf{е)}\ \frac{3^{8} \cdot 5^{8}}{3^{10} \cdot 5^{7}} = \frac{5^{8 - 7}}{3^{10 - 8}} = \frac{5}{3^{2}} = \frac{5}{9}\]