Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 284

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\[1)\log_{3}\left( 1 - x^{3} \right)\]

\[1 - x^{3} > 0\]

\[- x^{3} > - 1\]

\[x^{3} < 1\]

\[x < 1\]

\[Ответ:\ \ x < 1.\]

\[2)\log_{2}\left( x^{3} + 8 \right)\]

\[x^{3} + 8 > 0\]

\[x^{3} > - 8\ \]

\[x > - 2\]

\[Ответ:\ \ x > - 2.\]

\[3)\log_{\frac{1}{4}}\left( x^{3} + x^{2} - 6x \right)\]

\[x^{3} + x^{2} - 6x > 0\]

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

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

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

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

\[- 3 < x < 0;\text{\ \ }x > 2\]

\(Ответ:\ \ - 3 < x < 0;\ \ x > 2.\)

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

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

\[x\left( x^{2} + x - 2 \right) > 0\]

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

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

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

\[- 2 < x < 0\ \ и\ \ x > 1\]

\[Ответ:\ \ - 2 < x < 0;\ \ x > 1.\]