Решебник по алгебре 7 класс Мерзляк ФГОС Задание 319

 

\[\boxed{\text{319\ (319).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ \left( 2a^{2} - 5a + 7 \right) + \ *\ = 5\]

\[* = 5 - \left( 2a^{2} - 5a + 7 \right)\]

\[* = 5 - 2a^{2} + 5a - 7\]

\[* = - 2a^{2} + 5a - 2.\]

\[2)\ \left( 2a^{2} - 5a + 7 \right) + \ * = 0\]

\[* = 0 - \left( 2a^{2} - 5a + 7 \right)\]

\[*\ = 0 - 2a^{2} + 5a - 7\]

\[*\ = - 2a^{2} + 5a - 7.\]

\[3)\ \left( 2a^{2} - 5a + 7 \right) + \ *\ = a^{2}\]

\[*\ = a^{2} - (2a^{2} - 5a + 7)\]

\[*\ = a^{2} - 2a^{2} + 5a - 7\]

\[*\ = - a^{2} + 5a - 7.\]

\[4)\ \left( 2a^{2} - 5a + 7 \right) + \ *\ = - 2a\]

\[*\ = - 2a - \left( 2a^{2} - 5a + 7 \right)\]

\[*\ = \ - 2a - 2a^{2} + 5a - 7\]

\[* = - 2a^{2} + 3a - 7\text{.\ }\]

\[\boxed{\text{319.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left( 6^{4} \right)^{4}\ :\left( 6^{5} \right)^{3} = 6^{16}\ :6^{15} =\]

\[= 6^{1} = 6\]

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

\[= 2^{9}\ :2^{8} = 2^{1} = 2\]

\[3)\ \frac{7^{14} \cdot \left( 7^{2} \right)^{3}}{\left( 7^{3} \right)^{6} \cdot 7^{2}} = \frac{7^{14} \cdot 7^{6}}{7^{18} \cdot 7^{2}} = \frac{7^{20}}{7^{20}} =\]

\[= 1\]

\[4)\ \frac{25^{3} \cdot 125^{2}}{5^{10}} = \frac{\left( 5^{2} \right)^{3} \cdot \left( 5^{3} \right)^{2}}{5^{10}} =\]

\[= \frac{5^{6} \cdot 5^{6}}{5^{10}} = \frac{5^{12}}{5^{10}} = 5^{2} = 25\]

\[5)\ \frac{3^{8} \cdot 7^{8}}{21^{7}} = \frac{(3 \cdot 7)^{8}}{21^{7}} = \frac{21^{8}}{21^{7}} =\]

\[= 21\]

\[6)\ \frac{5^{9} \cdot 4^{6}}{20^{6}} = \frac{5^{3} \cdot 5^{6} \cdot 4^{6}}{20^{6}} =\]

\[= \frac{5^{3} \cdot 20^{6}}{20^{6}} = 5^{3} = 125\]