Решебник по алгебре 7 класс Мерзляк ФГОС Задание 928

 

\[\boxed{\text{928\ (928).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ x² + y² = 0\]

\[x = 0,\ \ y = 0\]

\[Ответ:(0;0).\]

\[2)\ (x + 2)^{2} + (y - 3)^{2} = 0\]

\[x + 2 = 0\ \ \ \ \ \ \ \ \ \ y - 3 = 0\]

\[x = - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = 3\]

\[Ответ:( - 2;3).\]

\[3)\ x^{4} + y^{6} = - 4\]

\[x^{4} \geq 0;\ \ \ \ y^{6} \geq 0\]

\[Ответ:нет\ корней.\]

\[\boxed{\text{928.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{\left( 7^{4} \right)^{2}}{7^{6} \cdot 49} = \frac{7^{8}}{7^{6} \cdot 7^{2}} = \frac{7^{8}}{7^{8}} = 1\]

\[2)\ \frac{\left( 5^{5} \right)^{3}}{5^{10} \cdot 125} = \frac{5^{15}}{5^{10} \cdot 5^{3}} =\]

\[= \frac{5^{15}}{5^{13}} = 5^{2} = 25\]

\[3)\ \frac{42^{8}}{36^{3} \cdot 7^{8}} = \frac{6^{8} \cdot 7^{8}}{\left( 6^{2} \right)^{3} \cdot 7^{8}} =\]

\[= \frac{6^{8}}{6^{6}} = 6^{2} = 32\]

\[4)\ \frac{50^{3}}{4^{3} \cdot 5^{6}} = \frac{2^{3} \cdot 25^{3}}{\left( 2^{2} \right)^{3} \cdot 5^{6}} =\]

\[= \frac{2^{3} \cdot \left( 5^{2} \right)^{3}}{2^{6} \cdot 5^{6}} = \frac{5^{6}}{2^{3} \cdot 5^{6}} = \frac{1}{8}\]

\[5)\ \frac{\left( 3^{20} + 3^{18} - 2 \cdot 3^{19} \right)}{2^{3} \cdot 9^{9}} =\]

\[= \frac{\left( 3^{10} - 3^{9} \right)^{2}}{2^{3} \cdot \left( 3^{2} \right)^{9}} = \frac{({3^{2})}^{9}(3 - 1)^{2}}{2^{3} \cdot 3^{18}} =\]

\[= \frac{3^{18} \cdot 2^{2}}{2^{3} \cdot 3^{18}} = \frac{1}{2} = 0,5\]

\[6)\ \frac{2^{48} - 2^{47} + 15 \cdot 2^{46}}{17 \cdot 16^{11}} =\]

\[= \frac{2^{46}\left( 2^{2} - 2 + 15 \right)}{17 \cdot \left( 2^{4} \right)^{11}} =\]

\[= \frac{2^{46} \cdot 17}{17 \cdot 2^{44}} = 2^{2} = 4\]