Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 350

\[\boxed{\mathbf{350}\mathbf{.}}\]

\[1)\lg\left( 6 \bullet 5^{x} - 25 \bullet 20^{x} \right) - \lg 25 =\]

\[= x\]

\[\lg\frac{6 \bullet 5^{x} - 25 \bullet 20^{x}}{25} = \lg 10^{x}\]

\[\frac{6 \bullet 5^{x} - 25 \bullet 20^{x}}{25} = 10^{x}\]

\[6 \bullet 5^{x} - 25 \bullet 20^{x} = 25 \bullet 10^{x}\]

\[25 \bullet 2^{x} + 25 \bullet 4^{x} - 6 = 0\]

\[25 \bullet 2^{2x} + 25 \bullet 2^{x} - 6 = 0\]

\[Пусть\ y = 2^{x}:\]

\[25y^{2} + 25y - 6 = 0\]

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

\[= 625 + 600 = 1225\]

\[y_{1} = \frac{- 25 - 35}{2 \bullet 25} = - \frac{60}{50} = - \frac{6}{5};\]

\[y_{2} = \frac{- 25 + 35}{2 \bullet 25} = \frac{10}{50} = \frac{1}{5}.\]

\[1)\ 2^{x} = - \frac{6}{5}\]

\[нет\ корней.\]

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

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

\[\log_{2}2^{x} = \log_{2}5^{- 1}\]

\[x = - \log_{2}5.\]

\[Ответ:\ \ x = - \log_{2}5.\]

\[2)\lg\left( 2^{x} + x + 4 \right) = x - x\lg 5\]

\[\lg\left( 2^{x} + x + 4 \right) = \lg 10^{x} - \lg 5^{x}\]

\[\lg\left( 2^{x} + x + 4 \right) = \lg\left( \frac{10}{5} \right)^{x}\]

\[2^{x} + x + 4 = \left( \frac{10}{5} \right)^{x}\]

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

\[x + 4 = 0\ \]

\[x = - 4.\]

\[Ответ:\ \ x = - 4.\]