Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1407

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\[1)\ 3^{\log_{2}\frac{x - 1}{x + 2}} < \frac{1}{9}\]

\[3^{\log_{2}\frac{x - 1}{x + 2}} < 3^{- 2}\]

\[\log_{2}\frac{x - 1}{x + 2} < - 2\]

\[\log_{2}\frac{x - 1}{x + 2} < \log_{2}2^{- 2}\]

\[\frac{x - 1}{x + 2} < \frac{1}{4}\ \ \ \ \ | \bullet {4(x + 2)}^{2}\]

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

\[4\left( x^{2} + 2x - x - 2 \right) < x^{2} + 4x + 4\]

\[4x^{2} + 4x - 8 < x^{2} + 4x + 4\]

\[3x^{2} < 12\]

\[x^{2} < 4\]

\[- 2 < x < 2.\]

\[Имеет\ смысл\ при:\]

\[\frac{x - 1}{x + 2} > 0\]

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

\[x < - 2\ \ и\ \ x > 1.\]

\[Ответ:\ \ 1 < x < 2.\]

\[2)\ 5^{\log_{2}\left( x^{2} - 4x + 3,5 \right)} > \frac{1}{5}\]

\[5^{\log_{2}\left( x^{2} - 4x + 3,5 \right)} > 5^{- 1}\]

\[\log_{2}\left( x^{2} - 4x + 3,5 \right) > - 1\]

\[\log_{2}\left( x^{2} - 4x + 3,5 \right) > \log_{2}2^{- 1}\]

\[x^{2} - 4x + 3,5 > 0,5\]

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

\[D = 16 - 12 = 4\]

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

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

\[(x - 1)(x - 3) > 0\]

\[x < 1\ \ и\ \ x > 3.\]

\[Ответ:\ \ x < 1;\ \ x > 3.\]