ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 259

\[\boxed{\mathbf{259}.}\]

\[1)\ 6^{12} - 1 = \left( 6^{2} \right)^{6} - 1 =\]

\[= (37 - 1)^{6} - 1 =\]

\[= 37^{6} - 6 \cdot 37^{5} + 15 \cdot 37^{4} -\]

\[- 20 \cdot 37^{3} + 15 \cdot 37^{2} -\]

\[- 6 \cdot 37 + \underset{0}{\overset{1^{6} - 1}{︸}}\ \vdots 37.\]

\[2)\ 2^{48} - 1 = \left( 2^{6} \right)^{8} - 1 =\]

\[= (65 - 1)^{8} - 1 =\]

\[= 65^{8} - 8 \cdot 65^{7} + 28 \cdot 65^{6} -\]

\[- 56 \cdot 65^{5} + 70 \cdot 65^{4} -\]

\[- 56 \cdot 65^{3} + 28 \cdot 65^{2} -\]

\[- 8 \cdot 65 + \underset{0}{\overset{1^{8} - 1}{︸}}\ \vdots 65.\]

\[3)\ 3^{17} - 3 = 3 \cdot \left( 3^{16} - 1 \right) =\]

\[= 3 \cdot \left( \left( 3^{4} \right)^{4} - 1 \right) =\]

\[= 3 \cdot (\left( 3^{4} \right)^{2} - 1)\left( 3^{8} + 1 \right) =\]

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

\[= \underset{240}{\overset{3 \cdot 80}{︸}} \cdot\]

\[\cdot \left( 3^{4} + 1 \right)\left( 3^{8} + 1 \right)\ \vdots 240.\]

\[4)\ 10^{12} + 263 = 10^{12} - 1 + 1 +\]

\[+ 263 = 10^{12} - 1 + 264;\]

\[так\ как\ 264\ :11\ без\ остатка,\ \]

\[надо\ доказать,\ \]

\[что\ \left( 10^{12} - 1 \right)\ \vdots 11.\]

\[10^{12} - 1 = (11 - 1)^{12} - 1 =\]

\[= 11^{12} - 12 \cdot 11^{11}\ldots - 12 \cdot 11 +\]

\[+ \underset{0}{\overset{1^{2} - 1\ }{︸}} \vdots 11.\]

\[5)\ 10^{24} - 298 = 10^{24} - 100 -\]

\[- 198;так\ как\ 198\ \vdots 99,\ надо\ \]

\[доказать,\ что\]

\[\left( 10^{21} - 100 \right)\ \vdots 99.\]

\[10^{24} - 100 = 100 \cdot \left( 10^{22} - 1 \right)\]

\[10^{22} - 1 = 100^{11} - 1 =\]

\[= (99 + 1)^{11} - 1 =\]

\[= 99^{11} + 55 \cdot 99^{10} + \ldots +\]

\[+ 11 \cdot 99 + \underset{0}{\overset{1^{11} - 1\ }{︸}} \vdots 99.\]