Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1352

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

\[\lg\left( \frac{1 + 2x}{2} \right) = \lg\frac{1}{2x}\]

\[\frac{1 + 2x}{2} = \frac{1}{2x}\ \ \ \ \ | \bullet 2x\]

\[x(1 + 2x) = 1\]

\[2x^{2} + x - 1 = 0\]

\[D = 1 + 8 = 9\]

\[x_{1} = \frac{- 1 - 3}{2 \bullet 2} = - 1;\ \]

\[x_{2} = \frac{- 1 + 3}{2 \bullet 2} = \frac{1}{2}.\]

\[Имеет\ смысл:\]

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

\[x > - \frac{1}{2};\]

\[x > 0.\]

\[Ответ:\ \ x = \frac{1}{2}.\]

\[2)\ 2\lg x = - \lg\frac{1}{6 - x^{2}}\]

\[\lg x^{2} = \lg\left( 6 - x^{2} \right)\]

\[x^{2} = 6 - x^{2}\]

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

\[x^{2} = 3\]

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

\[Имеет\ смысл:\]

\[6 - x^{2} > 0\]

\[- \sqrt{6} < x < \sqrt{6};\]

\[x > 0.\]

\[Ответ:\ \ x = \sqrt{3}.\]