Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 252

\[\boxed{\mathbf{252}\mathbf{.}}\]

\[1)\ 5^{2x} - 5^{x} - 600 = 0\ \]

\[Пусть\ y = 5^{x}:\]

\[y^{2} - y - 600 = 0\ \]

\[D = 1^{2} + 4 \bullet 600 = 1 + 1200 =\]

\[= 1201\]

\[y_{1} = \frac{1 - 49}{2} = - 24;\]

\[\text{\ \ }y_{2} = \frac{1 + 49}{2} = 25.\ \]

\[1)\ 5^{x} = - 24\]

\[нет\ корней.\ \]

\[2)\ 5^{x} = 25\ \]

\[5^{x} = 5^{2}\]

\[x = 2.\]

\[Ответ:\ \ x = 2.\]

\[2)\ 9^{x} - 3^{x} - 6 = 0\ \]

\[3^{2x} - 3^{x} - 6 = 0\ \]

\[Пусть\ y = 3^{x}:\]

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

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

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

\[y_{2} = \frac{1 + 5}{2} = 3.\ \]

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

\[нет\ корней.\ \]

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

\[x = 1.\ \]

\[Ответ:\ \ x = 1.\]

\[3)\ 3^{x} + 9^{x - 1} - 810 = 0\ \]

\[3^{x} + 9^{x} \bullet \frac{1}{9} - 810 = 0\ \ \ \ \ | \bullet 9\ \]

\[3^{2x} + 9 \bullet 3^{x} - 7290 = 0\ \]

\[Пусть\ y = 3^{x}:\]

\[y^{2} + 9y - 7290 = 0\ \]

\[D = 9^{2} + 4 \bullet 7290 =\]

\[= 81 + 29\ 160 = 29\ 241\]

\[y_{1} = \frac{- 9 - 171}{2} = - 90;\text{\ \ }\]

\[y_{2} = \frac{- 9 + 171}{2} = 81.\ \]

\[1)\ 3^{x} = - 90\]

\[нет\ корней.\ \]

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

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

\[x = 4.\ \]

\[Ответ:\ \ x = 4.\]

\[4)\ 4^{x} + 2^{x + 1} - 80 = 0\ \]

\[2^{2x} + 2 \bullet 2^{x} - 80 = 0\ \]

\[Пусть\ y = 2^{x}:\]

\[y^{2} + 2y - 80 = 0\ \]

\[D = 2^{2} + 4 \bullet 80 = 4 + 320 = 324\]

\[y_{1} = \frac{- 2 - 18}{2} = - 10;\text{\ \ }\]

\[y_{2} = \frac{- 2 + 18}{2} = 8.\ \]

\[1)\ 2^{x} = - 10\]

\[нет\ корней.\ \]

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

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

\[x = 3.\ \]

\[Ответ:\ \ x = 3.\]