Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 352

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

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

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

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

\[Пусть\ y = \log_{x}5:\]

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

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

\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]

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

\[y_{2} = \frac{2 + 4}{2} = 3.\]

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

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

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

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

\[x = \frac{1}{5} = 0,2.\]

\[2)\ \log_{x}5 = 3\]

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

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

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

\[Проверка:\]

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

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

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

\[= \sqrt{6 + 3} - 1\ :\frac{1}{3} = \sqrt{9} - 3 =\]

\[= 3 - 3 = 0.\]

\[Ответ:\ \ x = \sqrt[3]{5}.\]

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

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

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

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

\[Пусть\ y = \log_{2}x:\]

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

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

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

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

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

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

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

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

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

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

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

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

\[x = 2^{2}\]

\[x = 4.\]

\[Проверка:\]

\[= \sqrt{18 - 9 - 5} + 2 = \sqrt{4} + 2 =\]

\[= 2 + 2 = 4;\]

\[\sqrt{2\log_{2}^{2}4 + 3\log_{2}4 - 5} - \log_{2}8 =\]

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

\[= \sqrt{8 + 6 - 5} - 3 =\]

\[= \sqrt{9} - 3 = 3 - 3 = 0.\]

\[Ответ:\ \ x = 4.\]