Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 211

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\[1)\ 3^{2x - 1} + 3^{2x} = 108\]

\[3^{2x} \bullet \left( 3^{- 1} + 1 \right) = 108\]

\[3^{2x} \bullet \left( \frac{1}{3} + \frac{3}{3} \right) = 108\]

\[3^{2x} \bullet \frac{4}{3} = 108\]

\[3^{2x} = 81\]

\[3^{2x} = 3^{4}\]

\[2x = 4\]

\[x = 2\]

\[Ответ:\ \ x = 2.\]

\[2)\ 2^{3x + 2} - 2^{3x - 2} = 30\]

\[2^{3x} \bullet \left( 2^{2} - 2^{- 2} \right) = 30\]

\[2^{3x} \bullet \left( 4 - \frac{1}{2^{2}} \right) = 30\]

\[2^{3x} \bullet \left( \frac{16}{4} - \frac{1}{4} \right) = 30\]

\[2^{3x} \bullet \frac{15}{4} = 30\]

\[2^{3x} = 8\]

\[2^{3x} = 2^{3}\]

\[3x = 3\]

\[x = 1\]

\[Ответ:\ \ x = 1.\]

\[3)\ 2^{x + 1} + 2^{x - 1} + 2^{x} = 28\]

\[2^{x} \bullet \left( 2^{1} + 2^{- 1} + 1 \right) = 28\]

\[2^{x} \bullet \left( 3 + \frac{1}{2} \right) = 28\]

\[2^{x} \bullet \left( \frac{6}{2} + \frac{1}{2} \right) = 28\]

\[2^{x} \bullet \frac{7}{2} = 28\]

\[2^{x} = 8\]

\[2^{x} = 2^{3}\]

\[x = 3\]

\[Ответ:\ \ x = 3.\]

\[4)\ 3^{x - 1} - 3^{x} + 3^{x + 1} = 63\]

\[3^{x} \bullet \left( 3^{- 1} - 1 + 3^{1} \right) = 63\]

\[3^{x} \bullet \left( \frac{1}{3} + 2 \right) = 63\]

\[3^{x} \bullet \left( \frac{1}{3} + \frac{6}{3} \right) = 63\]

\[3^{x} \bullet \frac{7}{3} = 63\]

\[3^{x} = 27\]

\[3^{x} = 3^{3}\]

\[x = 3\]

\[Ответ:\ \ x = 3.\]