ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1612

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\[1)\ \left( \frac{2}{5} \right)^{x^{2} - 5x + 6} < 1\]

\[\left( \frac{2}{5} \right)^{x^{2} - 5x + 6} < \left( \frac{2}{5} \right)^{0}\]

\[x^{2} - 5x + 6 > 0\]

\[D = 25 - 24 = 1\]

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

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

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

\[x < 2\ \ и\ \ x > 3.\]

\[Ответ:\ \ x \in ( - \infty;\ 2) \cup (3;\ + \infty).\]

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

\[5^{x} - 3 \bullet 3^{x} > 2 \bullet \left( \frac{5^{x}}{5} - \frac{3^{x}}{3^{2}} \right)\]

\[5^{x} - 3 \bullet 3^{x} - \frac{2}{5} \bullet 5^{x} + \frac{2}{9} \bullet 3^{x} > 0\]

\[\frac{3}{5} \bullet 5^{x} - \frac{25}{9} \bullet 3^{x} > 0\ \ \ \ \ | \bullet 45\]

\[27 \bullet 5^{x} - 125 \bullet 3^{x} > 0\]

\[3^{3} \bullet 5^{x} - 5^{3} \bullet 3^{x} > 0\]

\[3^{3} \bullet 5^{x} > 5^{3} \bullet 3^{x}\]

\[\left( \frac{3}{5} \right)^{3} > \left( \frac{3}{5} \right)^{x}\]

\[3 < x.\]

\[Ответ:\ \ x \in (3;\ + \infty).\]