Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 250

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\[1)\ {1,5}^{5x - 7} = \left( \frac{2}{3} \right)^{x + 1}\]

\[\left( \frac{15}{10} \right)^{5x - 7} = \left( \frac{3}{2} \right)^{- (x + 1)}\]

\[\left( \frac{3}{2} \right)^{5x - 7} = \left( \frac{3}{2} \right)^{- x - 1}\]

\[5x - 7 = - x - 1\]

\[6x = 6\]

\[x = 1\]

\[Ответ:\ \ x = 1.\]

\[2)\ {0,75}^{2x - 3} = \left( 1\frac{1}{3} \right)^{5 - x}\]

\[\left( \frac{75}{100} \right)^{2x - 3} = \left( \frac{1 \bullet 3 + 1}{3} \right)^{5 - x}\]

\[\left( \frac{3}{4} \right)^{2x - 3} = \left( \frac{4}{3} \right)^{5 - x}\]

\[\left( \frac{3}{4} \right)^{2x - 3} = \left( \frac{3}{4} \right)^{x - 5}\]

\[2x - 3 = x - 5\]

\[x = - 5 + 3\]

\[x = - 2\]

\[Ответ:\ \ x = - 2.\]

\[3)\ 5^{x^{2} - 5x - 6} = 1\]

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

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

\[D = 5^{2} + 4 \bullet 6 = 25 + 24 = 49\]

\[x_{1} = \frac{5 - 7}{2} = - 1;\ \text{\ \ }\]

\[x_{2} = \frac{5 + 7}{2} = 6.\]

\[Ответ:\ \ x_{1} = - 1;\ \ \ x_{2} = 6.\]

\[4)\ \left( \frac{1}{7} \right)^{x^{2} - 2x - 2} = \frac{1}{7}\]

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

\[x^{2} - 2x - 3 = 0\]

\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]

\[x_{1} = \frac{2 - 4}{2} = - 1;\ \text{\ \ }\]

\[x_{2} = \frac{2 + 4}{2} = 3.\]

\[Ответ:\ \ x_{1} = - 1;\ \ \ x_{2} = 3.\]