Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 222

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\[1)\ 3^{x + 3} + 3^{x} = 7^{x + 1} + 5 \bullet 7^{x}\]

\[3^{x} \bullet \left( 3^{3} + 1 \right) = 7^{x} \bullet \left( 7^{1} + 5 \right)\]

\[3^{x} \bullet (27 + 1) = 7^{x} \bullet (7 + 5)\]

\[3^{x} \bullet 28 = 7^{x} \bullet 12\]

\[\frac{3^{x}}{7^{x}} = \frac{12}{28}\]

\[\left( \frac{3}{7} \right)^{x} = \frac{3}{7}\]

\[x = 1\]

\[Ответ:\ \ x = 1.\]

\[2)\ 3^{x + 4} + 3 \bullet 5^{x + 3} = 5^{x + 4} + 3^{x + 3}\]

\[3^{x + 4} - 3^{x + 3} = 5^{x + 4} - 3 \bullet 5^{x + 3}\]

\[3^{x} \bullet \left( 3^{4} - 3^{3} \right) = 5^{x} \bullet \left( 5^{4} - 3 \bullet 5^{3} \right)\]

\[3^{x} \bullet (81 - 27) =\]

\[= 5^{x} \bullet (625 - 3 \bullet 125)\]

\[3^{x} \bullet 54 = 5^{x} \bullet 250\]

\[\frac{3^{x}}{5^{x}} = \frac{250}{54}\]

\[\left( \frac{3}{5} \right)^{x} = \frac{125}{27}\]

\[\left( \frac{3}{5} \right)^{x} = \left( \frac{5}{3} \right)^{3}\]

\[\left( \frac{3}{5} \right)^{x} = \left( \frac{3}{5} \right)^{- 3}\]

\[x = - 3\]

\[Ответ:\ \ x = - 3.\]

\[3)\ 2^{8 - x} + 7^{3 - x} =\]

\[= 7^{4 - x} + 2^{3 - x} \bullet 11\]

\[2^{8 - x} - 11 \bullet 2^{3 - x} = 7^{4 - x} - 7^{3 - x}\]

\[2^{- x} \bullet \left( 2^{8} - 11 \bullet 2^{3} \right) =\]

\[= 7^{- x} \bullet \left( 7^{4} - 7^{3} \right)\]

\[2^{- x} \bullet (256 - 11 \bullet 8) =\]

\[= 7^{- x} \bullet (2401 - 343)\]

\[2^{- x} \bullet 168 = 7^{- x} \bullet \ 2058\]

\[\frac{2^{- x}}{7^{- x}} = \frac{2058}{168}\]

\[\left( \frac{2}{7} \right)^{- x} = \frac{49}{4}\]

\[\left( \frac{7}{2} \right)^{x} = \left( \frac{7}{2} \right)^{2}\]

\[x = 2\]

\[Ответ:\ \ x = 2.\]

\[4)\ 2^{x + 1} + 2^{x - 1} - 3^{x - 1} =\]

\[= 3^{x - 2} - 2^{x - 3} + 2 \bullet 3^{x - 3}\]

\[2^{x + 1} + 2^{x - 1} + 2^{x - 3} =\]

\[= 3^{x - 2} + 3^{x - 1} + 2 \bullet 3^{x - 3}\]

\[2^{x} \bullet \left( 2^{1} + 2^{- 1} + 2^{- 3} \right) =\]

\[= 3^{x} \bullet \left( 3^{- 2} + 3^{- 1} + 2 \bullet 3^{- 3} \right)\]

\[2^{x} \bullet \left( 2 + \frac{1}{2} + \frac{1}{8} \right) =\]

\[= 3^{x} \bullet \left( \frac{1}{9} + \frac{1}{3} + \frac{2}{27} \right)\]

\[2^{x} \bullet \left( \frac{16}{8} + \frac{4}{8} + \frac{1}{8} \right) =\]

\[= 3^{x} \bullet \left( \frac{3}{27} + \frac{9}{27} + \frac{2}{27} \right)\]

\[2^{x} \bullet \frac{21}{8} = 3^{x} \bullet \frac{14}{27}\]

\[\frac{2^{x}}{3^{x}} = \frac{14}{27} \bullet \frac{8}{21}\]

\[\left( \frac{2}{3} \right)^{x} = \frac{16}{81}\]

\[\left( \frac{2}{3} \right)^{x} = \left( \frac{2}{3} \right)^{4}\]

\[x = 4\]

\[Ответ:\ \ x = 4.\]