Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 855

\[1)\log_{3}{3x} + \log_{3}(4x + 1) = \log_{4x^{2} + x}9\]

\[\log_{3}3 + \log_{3}x + \log_{3}(4x + 1) =\]

\[= \frac{\log_{3}9}{\log_{3}\left( 4x^{2} + x \right)}\]

\[1 + \log_{3}\left( 4x^{2} + x \right) = \frac{2}{\log_{3}\left( 4x^{2} + x \right)}\]

\[y = \log_{3}\left( 4x^{2} + x \right):\]

\[1 + y = \frac{2}{y}\ \ \ \ \ | \bullet y\]

\[y + y^{2} = 2\]

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

\[D = 1 + 8 = 9\]

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

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

\[1)\ \log_{3}\left( 4x^{2} + x \right) = - 2\]

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

\[4x^{2} + x = \frac{1}{9}\]

\[36x^{2} + 9x - 1 = 0\]

\[D = 81 + 144 = 225\]

\[x_{1} = \frac{- 9 - 15}{2 \bullet 36} = - \frac{1}{3};\ \]

\[x_{2} = \frac{- 9 + 15}{2 \bullet 36} = \frac{1}{12}.\]

\[2)\ \log_{3}\left( 4x^{2} + x \right) = 1\]

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

\[4x^{2} + x - 3 = 0\]

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{- 1 - 7}{2 \bullet 4} = - 1;\]

\[x_{2} = \frac{- 1 + 7}{2 \bullet 4} = \frac{3}{4}.\]

\[Область\ определения:\]

\[4x^{2} + x > 0\]

\[4x\left( x + \frac{1}{4} \right) > 0\]

\[x > 0;\ \ \ x > - \frac{1}{4}.\]

\[4x^{2} + x \neq 1.\]

\[Ответ:\ \ \frac{1}{12};\ \frac{3}{4}.\]

\[2)\log_{2}\frac{x}{2} + \log_{2}(21x - 2) =\]

\[= 2\log_{21x^{2} - 2x}8\]

\[\log_{2}x - \log_{2}2 + \log_{2}(21x - 2) =\]

\[= 2 \bullet \frac{\log_{2}8}{\log_{2}\left( 21x^{2} - 2x \right)}\]

\[\log_{2}\left( 21x^{2} - 2x \right) - 1 =\]

\[= \frac{2 \bullet 3}{\log_{2}\left( 21x^{2} - 2x \right)}\]

\[y = \log_{2}\left( 21x^{2} - 2x \right):\]

\[y - 1 = \frac{6}{y}\ \ \ \ \ | \bullet y\]

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

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

\[D = 1 + 24 = 25\]

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

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

\[1)\ \log_{2}\left( 21x^{2} - 2x \right) = - 2\]

\[21x^{2} - 2x = 2^{- 2}\]

\[21x^{2} - 2x = \frac{1}{4}\]

\[84x^{2} - 8x - 1 = 0\]

\[D = 64 + 336 = 400\]

\[x_{1} = \frac{8 - 20}{2 \bullet 84} = - \frac{1}{14};\]

\[x_{2} = \frac{8 + 20}{2 \bullet 84} = \frac{1}{6}.\]

\[2)\ \log_{2}\left( 21x^{2} - 2x \right) = 3\]

\[21x^{2} - 2x = 2^{3}\]

\[21x^{2} - 2x - 8 = 0\]

\[D = 4 + 672 = 676\]

\[x_{1} = \frac{2 - 26}{2 \bullet 21} = - \frac{4}{7};\]

\[x_{2} = \frac{2 + 26}{2 \bullet 21} = \frac{2}{3}.\]

\[Область\ определения:\]

\[21x^{2} - 2x > 0\]

\[21x\left( x - \frac{2}{21} \right) > 0\text{\ \ }\]

\[x > 0;\ \ \ x > \frac{2}{21}.\]

\[21x^{2} - 2x \neq 1.\]

\[Ответ:\ \ \frac{1}{6};\ \frac{2}{3}.\]