Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 855

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\[1)\ \sqrt{\log_{x}25 + 3} = \frac{1}{\log_{5}x}\ \]

\[\sqrt{\log_{x}25 + 3} = \log_{x}5\]

\[t = \log_{x}5:\]

\[\sqrt{2t + 3} = t\]

\[2t + 3 = t^{2}\]

\[t^{2} - 2t - 3 = 0\]

\[t_{1} = 3;\ \ t_{2} = - 1.\]

\[\textbf{а)}\ \log_{x}5 = 3\]

\[\log_{x}5 = \log_{x}x^{3}\]

\[x^{3} = 5\]

\[x = \sqrt[3]{5}.\]

\[\textbf{б)}\ \log_{x}5 = - 1\]

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

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

\[\frac{1}{x} = 5\]

\[x = \frac{1}{5}\]

\[Проверка\ показывает,\ что\ x =\]

\[= \frac{1}{5}\ не\ является\ корнем\]

\[\ уравнения.\]

\[Ответ:x = \sqrt[3]{5}.\]

\[2)\ \sqrt{2\log_{2}^{2}x + 3\log_{2}x - 5} =\]

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

\[\sqrt{2\log_{2}^{2}x + 3\log_{2}x - 5} =\]

\[= \log_{2}2 + \log_{2}x\]

\[Пусть\ y = \log_{2}x:\]

\[\sqrt{2y^{2} + 3y - 5} = 1 + y\]

\[2y^{2} + 3y - 5 = 1 + 2y + y^{2}\]

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

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

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

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

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

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

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

\[x = \frac{1}{8}.\]

\[2)\ \log_{2}x = 2\]

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

\[x = 2^{2}\]

\[x = 4.\]

\[Проверка:\]

\[\sqrt{2\log_{2}^{2}\frac{1}{8} + 3\log_{2}\frac{1}{8} - 5} -\]