Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 223

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\[1)\ 8 \bullet 4^{x} - 6 \bullet 2^{x} + 1 = 0\]

\[8 \bullet 2^{2x} - 6 \bullet 2^{x} + 1 = 0\]

\(Пусть\ y = 2^{x}:\)

\[8y^{2} - 6y + 1 = 0\ \]

\[D = 6^{2} - 4 \bullet 8 = 36 - 32 = 4\]

\[y_{1} = \frac{6 - 2}{2 \bullet 8} = \frac{4}{16} = \frac{1}{4};\]

\[y_{2} = \frac{6 + 2}{2 \bullet 8} = \frac{8}{16} = \frac{1}{2}.\]

\[1)\ 2^{x} = \frac{1}{4}\]

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

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

\[x = - 2.\]

\[2)\ 2^{x} = \frac{1}{2}\]

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

\[x = - 1.\]

\[Ответ:\ \ x_{1} = - 2;\ \ \ x_{2} = - 1.\]

\[2)\ \left( \frac{1}{4} \right)^{x} + \left( \frac{1}{2} \right)^{x} - 6 = 0\]

\[\left( \frac{1}{2} \right)^{2x} + \left( \frac{1}{2} \right)^{x} - 6 = 0\]

\[Пусть\ y = \left( \frac{1}{2} \right)^{x}:\]

\[y^{2} + y - 6 = 0\]

\[D = 1^{2} + 4 \bullet 6 = 1 + 24 = 25\]

\[y_{1} = \frac{- 1 - 5}{2} = - 3;\ \text{\ \ }\]

\[y_{2} = \frac{- 1 + 5}{2} = 2.\]

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

\[нет\ корней.\]

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

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

\[x = - 1.\]

\[Ответ:\ \ x = - 1.\]

\[3)\ 13^{2x + 1} - 13^{x} - 12 = 0\]

\[13 \bullet 13^{2x} - 13^{x} - 12 = 0\]

\[Пусть\ y = 13^{x}:\]

\[13y^{2} - y - 12 = 0\]

\[D = 1^{2} + 4 \bullet 13 \bullet 12 =\]

\[= 1 + 624 = 625\]

\[y_{1} = \frac{1 - 25}{2 \bullet 13} = - \frac{24}{26} = - \frac{12}{13};\ \]

\[y_{2} = \frac{1 + 25}{2 \bullet 13} = \frac{26}{26} = 1.\]

\[1)\ 13^{x} = - \frac{12}{13}\]

\[нет\ корней.\]

\[2)\ 13^{x} = 1\]

\[13^{x} = 13^{0}\]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]

\[4)\ 3^{2x + 1} - 10 \bullet 3^{x} + 3 = 0\]

\[3 \bullet 3^{2x} - 10 \bullet 3^{x} + 3 = 0\]

\[Пусть\ y = 3^{x}:\]

\[3y^{2} - 10y + 3 = 0\]

\[D = 10^{2} - 4 \bullet 3 \bullet 3 =\]

\[= 100 - 36 = 64\]

\[y_{1} = \frac{10 - 8}{2 \bullet 3} = \frac{2}{6} = \frac{1}{3};\]

\[y_{2} = \frac{10 + 8}{2 \bullet 3} = \frac{18}{6} = 3.\]

\[1)\ 3^{x} = \frac{1}{3}\ \]

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

\[x = - 1.\]

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

\[3^{x} = 3^{1}\]

\[x = 1.\]

\[Ответ:\ \ x = \pm 1.\]

\[5)\ 2^{3x} + 8 \bullet 2^{x} - 6 \bullet 2^{2x} = 0\]

\[2^{x} \bullet \left( 2^{2x} + 8 - 6 \bullet 2^{x} \right) = 0\]

\[2^{2x} - 6 \bullet 2^{x} + 8 = 0\]

\[Пусть\ y = 2^{x}:\]

\[y^{2} - 6y + 8 = 0\]

\[D = 6^{2} - 4 \bullet 8 = 36 - 32 = 4\]

\[y_{1} = \frac{6 - 2}{2} = 2;\text{\ \ \ }y_{2} = \frac{6 + 2}{2} = 4.\]

\[1)\ 2^{x} = 2\]

\[2^{x} = 2^{1}\]

\[x = 1.\]

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

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

\[x = 2.\]

\[Ответ:\ \ x_{1} = 1;\ \ \ x_{2} = 2.\]

\[6)\ 5^{3x + 1} + 34 \bullet 5^{2x} - 7 \bullet 5^{x} = 0\]

\[5^{x} \bullet \left( 5^{2x + 1} + 34 \bullet 5^{x} - 7 \right) = 0\]

\[5 \bullet 5^{2x} + 34 \bullet 5^{x} - 7 = 0\]

\[Пусть\ y = 5^{x}:\]

\[5y^{2} + 34y - 7 = 0\]

\[D = 34^{2} + 4 \bullet 5 \bullet 7 =\]

\[= 1156 + 140 = 1296\]

\[y_{1} = \frac{- 34 - 36}{2 \bullet 5} = - \frac{70}{10} = - 7;\ \]

\[y_{2} = \frac{- 34 + 36}{2 \bullet 5} = \frac{2}{10} = \frac{1}{5}\]

\[1)\ 5^{x} = - 7\]

\[нет\ корней.\]

\[2)\ 5^{x} = \frac{1}{5}\]

\[5^{x} = 5^{- 1}\ \]

\[x = - 1.\]

\[Ответ:\ \ x = - 1.\]