Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1356

\[\boxed{\mathbf{1356}\mathbf{.}}\]

\[1)\ 7 \bullet 4^{x^{2}} - 9 \bullet 14^{x^{2}} + 2 \bullet 49^{x^{2}} = 0\]

\[7 \bullet 2^{2x^{2}} - 9 \bullet 7^{x^{2}} \bullet 2^{x^{2}} + 2 \bullet 7^{2x^{2}} = 0\ \ \ \ |\ :7^{2x^{2}}\]

\[7 \bullet \left( \frac{2}{7} \right)^{2x^{2}} - 9\left( \frac{2}{7} \right)^{x^{2}} + 2 = 0\]

\[y = \left( \frac{2}{7} \right)^{x^{2}}:\]

\[7y^{2} - 9y + 2 = 0\]

\[D = 81 - 56 = 25\]

\[y_{1} = \frac{9 - 5}{2 \bullet 7} = \frac{4}{14} = \frac{2}{7};\]

\[y_{2} = \frac{9 + 5}{2 \bullet 7} = 1.\]

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

\[x^{2} = 1\]

\[x = \pm 1.\]

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

\[x^{2} = 0\]

\[x = 0.\]

\[Ответ:\ \ x_{1} = \pm 1;\ \ x_{2} = 0.\ \]

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

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

\[625 \bullet 5^{x} + 192 \bullet 4^{x} = 256 \bullet 4^{x} + 500 \bullet 5^{x}\]

\[625 + 192 \bullet \left( \frac{4}{5} \right)^{x} = 256 \bullet \left( \frac{4}{5} \right)^{x} + 500\]

\[64 \bullet \left( \frac{4}{5} \right)^{x} = 125\]

\[\left( \frac{4}{5} \right)^{x} = \frac{125}{64}\]

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

\[- x = 3\]

\[x = - 3.\]

\[Ответ:\ \ x = - 3.\]