Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1353

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\[1)\log_{2}(2x - 18) + \log_{2}(x - 9) = 5\]

\[\log_{2}\left( (2x - 18)(x - 9) \right) = \log_{2}2^{5}\]

\[\log_{2}\left( 2x^{2} - 18x - 18x + 162 \right) = \log_{2}32\]

\[2x^{2} - 36x + 162 = 32\]

\[2x^{2} - 36x + 130 = 0\]

\[x^{2} - 18x + 65 = 0\]

\[D = 324 - 260 = 64\]

\[x_{1} = \frac{18 - 8}{2} = 5;\]

\[x_{2} = \frac{18 + 8}{2} = 13.\]

\[ОДЗ:\]

\[2x - 18 > 0\]

\[x > 9.\]

\[x - 9 > 0\]

\[x > 9.\]

\[Ответ:\ \ x = 13.\]

\[2)\lg\left( x^{2} + 19 \right) - \lg(x + 1) = 1\]

\[\lg\frac{x^{2} + 19}{x + 1} = \lg 10\]

\[\frac{x^{2} + 19}{x + 1} = 10\ \ \ \ \ | \bullet (x + 1)\ \]

\[x^{2} + 19 = 10(x + 1)\]

\[x^{2} + 19 = 10x + 10\]

\[x^{2} - 10x + 9 = 0\]

\[D = 100 - 36 = 64\]

\[x_{1} = \frac{10 - 8}{2} = 1;\]

\[x_{2} = \frac{10 + 8}{2} = 9.\]

\[ОДЗ:\]

\[x^{2} + 19 > 0\]

\[при\ любом\ \text{x.}\]

\[x + 1 > 0\]

\[x > - 1.\]

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