Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 344

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\[\log_{4}\frac{(x + 2)(x + 3)(x - 2)}{x + 3} =\]

\[= \log_{4}4^{2}\]

\[\log_{4}\left( x^{2} - 4 \right) = \log_{4}16\]

\[x^{2} - 4 = 16\]

\[x^{2} = 20\]

\[x = \pm \sqrt{20}\]

\[x = \pm 2\sqrt{5}.\]

\[имеет\ смысл\ при:\]

\[1)\ (x + 3)(x + 2) > 0\]

\[x < - 3;\text{\ \ }x > - 2.\]

\[2)\ \frac{x - 2}{x + 3} > 0\]

\[(x + 3)(x - 2) > 0\]

\[x < - 3;\text{\ \ }x > 2.\]

\[Ответ:\ \ x = \pm 2\sqrt{5}\]

\[\log_{2}\frac{(x - 1)(x - 1)(x + 4)}{x + 4} =\]

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

\[\log_{2}(x - 1)^{2} = \log_{2}4\]

\[(x - 1)^{2} = 4\]

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

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

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

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

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

\[имеет\ смысл\ при:\]

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

\[(x + 4)(x - 1) > 0\]

\[x < - 4;\ \text{\ \ }x > 1.\]

\[Ответ:\ \ x = 3.\]

\[3)\log_{3}x^{2} - \log_{3}\frac{x}{x + 6} = 3\]

\[\log_{3}\left( x^{2}\ :\frac{x}{x + 6} \right) = 3\]

\[\log_{3}\frac{x^{2} \bullet (x + 6)}{x} = \log_{3}3^{3}\]

\[\log_{3}\left( x^{2} + 6x \right) = \log_{3}27\]

\[x^{2} + 6x = 27\]

\[x^{2} + 6x - 27 = 0\]

\[D = 6^{2} + 4 \bullet 27 = 36 + 108 =\]

\[= 144\]

\[x_{1} = \frac{- 6 - 12}{2} = - 9;\ \]

\[x_{2} = \frac{- 6 + 12}{2} = 3.\]

\[имеет\ смысл\ при:\]

\[\frac{x}{x + 6} > 0\]

\[(x + 6) \bullet x > 0\]

\[x < - 6;\ \ \text{\ \ }x > 0.\]

\[Ответ:\ \ x_{1} = - 9;\ \ x_{2} = 3.\]

\[4)\log_{2}\frac{x + 4}{x} + \log_{2}x^{2} = 5\]

\[\log_{2}\frac{x^{2} \bullet (x + 4)}{x} = \log_{2}2^{5}\]

\[\log_{2}\left( x^{2} + 4x \right) = \log_{2}32\]

\[x^{2} + 4x = 32\]

\[x^{2} + 4x - 32 = 0\]

\[D = 4^{2} + 4 \bullet 32 = 16 + 128 =\]

\[= 144\]

\[x_{1} = \frac{- 4 - 12}{2} = - 8;\ \ \ \ \ \]

\[x_{2} = \frac{- 4 + 12}{2} = 4.\]

\[имеет\ смысл\ при:\]

\[\frac{x + 4}{x} > 0\]

\[(x + 4) \bullet x > 0\]

\[x < - 4;\text{\ \ }x > 0\]

\[Ответ:\ \ x_{1} = - 8;\ \ x_{2} = 4.\]