Решебник по алгебре 7 класс Мерзляк ФГОС Задание 320

 

\[\boxed{\text{320\ (320).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left( 4a^{3} - 8 \right) - \ *\ = - 4\]

\[*\ = 4a^{3} - 8 - ( - 4)\]

\[*\ = 4a^{3} - 8 + 4\]

\[*\ = 4a^{3} - 4.\]

\[2)\ (4a^{3} - 8) - \ *\ = 9\]

\[*\ = 4a^{3} - 8 - 9\]

\[* = 4a^{3} - 17.\]

\[3)\ \left( 4a^{3} - 8 \right) - \ *\ = - 2a^{3}\]

\[*\ = 4a^{3} - 8 - \left( - 2a^{3} \right)\]

\[*\ = 4a^{3} - 8 + 2a^{3}\]

\[*\ = 6a^{3} - 8.\]

\[4)\ (4a^{3} - 8) - \ *\ = 3a\]

\[*\ = 4a^{3} - 8 - 3a\]

\[*\ = 4a^{3} - 3a - 8\text{.\ }\]

\[\boxed{\text{320.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 100^{5}\ :1000^{2} =\]

\[= \left( 10^{2} \right)^{5}\ :\left( 10^{3} \right)^{2} = 10^{10\ }:10^{6} =\]

\[= 10^{4} = 10\ 000\]

\[2)\ \frac{3^{10} \cdot \left( 3^{3} \right)^{5}}{\left( 3^{5} \right)^{4} \cdot 3} = \frac{3^{10} \cdot 3^{15}}{3^{20} \cdot 3} =\]

\[= \frac{3^{25}}{3^{21}} = 3^{4} = 81\]

\[3)\ \frac{4^{3} \cdot 16^{2}}{2^{12}} = \frac{\left( 2^{2} \right)^{3} \cdot \left( 2^{4} \right)^{2}}{2^{12}} =\]

\[= \frac{2^{6} \cdot 2^{8}}{2^{12}} = \frac{2^{14}}{2^{12}} = 2^{2} = 4\]

\[4)\ \frac{45^{10}}{5^{8} \cdot 3^{19}} = \frac{\left( 5 \cdot 3^{2} \right)^{10}}{5^{8} \cdot 3^{19}} =\]

\[= \frac{5^{10} \cdot 3^{20}}{5^{8} \cdot 3^{19}} = 5^{2} \cdot 3 = 25 \cdot 3 =\]

\[= 75\]