Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 244

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\[1)\ \left\{ \begin{matrix} 5^{2x + 1} > 625\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 11^{6x^{2} - 10x} = 11^{9x - 15} \\ \end{matrix} \right.\ \]

\[1)\ 5^{2x + 1} > 625\]

\[5^{2x + 1} > 5^{4}\]

\[2x + 1 > 4\]

\[2x > 3\ \]

\[x > 1,5.\]

\[2)\ 11^{6x^{2} - 10x} = 11^{9x - 15}\]

\[6x^{2} - 10x = 9x - 15\]

\[6x^{2} - 19x + 15 = 0\]

\[D = 19^{2} - 4 \bullet 6 \bullet 15 =\]

\[= 361 - 360 = 1\]

\[x_{1} = \frac{19 - 1}{2 \bullet 6} = \frac{18}{12} = \frac{3}{2} = 1,5;\ \]

\[x_{2} = \frac{19 + 1}{2 \bullet 6} = \frac{20}{12} = \frac{5}{3} = 1\frac{2}{3}.\]

\[Ответ:\ \ x = 1\frac{2}{3}.\]

\[2)\ \left\{ \begin{matrix} {0,3}^{10x^{2} - 47x} = {0,3}^{- 10x - 7} \\ {3,7}^{x^{2}} = {3,7}^{0,04}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[1)\ {0,3}^{10x^{2} - 47x} = {0,3}^{- 10x - 7}\]

\[10x^{2} - 47x = - 10x - 7\]

\[10x^{2} - 37x + 7 = 0\]

\[D = 37^{2} - 4 \bullet 10 \bullet 7 =\]

\[= 1369 - 280 = 1089\]

\[x_{1} = \frac{37 - 33}{2 \bullet 10} = \frac{4}{20} = \frac{1}{5} = 0,2;\ \]

\[x_{2} = \frac{37 + 33}{2 \bullet 10} = \frac{70}{20} = \frac{7}{2} = 3,5.\]

\[2)\ {3,7}^{x^{2}} = {3,7}^{0,04}\]

\[x^{2} = 0,04\]

\[x = \pm 0,2.\]

\[Ответ:\ \ x = 0,2.\]