Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 741

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\[1)\ 2^{x + 4} + 2^{x + 2} = 5^{x + 1} + 3 \bullet 5^{x}\ \]

\[2^{x} \bullet \left( 2^{4} + 2^{2} \right) = 5^{x} \bullet \left( 5^{1} + 3 \right)\ \]

\[2^{x} \bullet (16 + 4) = 5^{x} \bullet 8\ \]

\[2^{x} \bullet 20 = 5^{x} \bullet 8\ \]

\[\frac{2^{x}}{5^{x}} = \frac{8}{20}\ \]

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

\[x = 1\ \]

\[Ответ:\ \ x = 1.\]

\[2)\ 5^{2x} - 7^{x} - 5^{2x} \bullet 17 +\]

\[+ 7^{x} \bullet 17 = 0\ \]

\[5^{2x} - 17 \bullet 5^{2x} = 7^{x} - 17 \bullet 7^{x}\ \]

\[5^{2x} \bullet (1 - 17) = 7^{x} \bullet (1 - 17)\ \]

\[5^{2x} = 7^{x}\ \]

\[2x = x\ \]

\[x = 0\ \]

\[Ответ:\ \ x = 0.\]

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

\[2^{x^{2} - 1} + 2^{x^{2} + 2} = 3^{x^{2} - 1} + 3^{x^{2}}\ \]

\[2^{x^{2}} \bullet \left( 2^{- 1} + 2^{2} \right) =\]

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

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

\[2^{x^{2}} \bullet \left( \frac{1}{2} + \frac{8}{2} \right) = 3^{x^{2}} \bullet \left( \frac{1}{3} + \frac{3}{3} \right)\ \]

\[2^{x^{2}} \bullet \frac{9}{2} = 3^{x^{2}} \bullet \frac{4}{3}\ \]

\[\left( \frac{2}{3} \right)^{x^{2}} = \frac{8}{27}\ \]

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

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

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

\[Ответ:\ \ x = \pm \sqrt{3}.\]

\[4)\ 3 \bullet 4^{x} + \frac{1}{3} \bullet 9^{x + 2} =\]

\[= 6 \bullet 4^{x + 1} - \frac{1}{2} \bullet 9^{x + 1}\ \]

\[3 \bullet 4^{x} - 6 \bullet 4^{x + 1} =\]

\[= - \frac{1}{2} \bullet 9^{x + 1} - \frac{1}{3} \bullet 9^{x + 2}\ \]

\[4^{x} \bullet \left( 3 - 6 \bullet 4^{1} \right) =\]

\[= 9^{x} \bullet \left( - \frac{1}{2} \bullet 9^{1} - \frac{1}{3} \bullet 9^{2} \right)\ \]

\[4^{x} \bullet (3 - 24) = 9^{x} \bullet \left( - \frac{9}{2} - \frac{81}{3} \right)\ \]

\[4^{x} \bullet ( - 21) = 9^{x} \bullet \left( - \frac{9}{2} - \frac{27 \bullet 2}{2} \right)\ \]

\[4^{x} \bullet ( - 21) = 9^{x} \bullet \left( - \frac{63}{2} \right)\ \]

\[\frac{4^{x}}{9^{x}} = \frac{3}{2}\ \]

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

\[2x = - 1\]

\[x = - 0,5\ \]

\[Ответ:\ \ x = - 0,5.\]