ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 715

\[\boxed{\mathbf{715}.}\]

\[(2,5)^{(x + 1)^{2}} \cdot (0,4)^{4(x - 1)} \geq \left( \frac{25}{4} \right)^{6,5}\]

\[\left( \frac{5}{2} \right)^{(x + 1)^{2}} \cdot \left( \frac{2}{5} \right)^{4 \cdot (x - 1)} \geq \left( \frac{5}{2} \right)^{13}\]

\[\left( \frac{5}{2} \right)^{(x + 1)^{2}} \cdot \left( \frac{5}{2} \right)^{- 4 \cdot (x - 1)} \geq \left( \frac{5}{2} \right)^{13}\]

\[\left( \frac{5}{2} \right)^{(x + 1)^{2} - 4 \cdot (x - 1)} \geq \left( \frac{5}{2} \right)^{13}\]

\[Так\ как\ функция\ \]

\[возрастающая:\]

\[(x + 1)^{2} - 4 \cdot (x - 1) \geq 13\]

\[при\ x = 1 - решений\ нет.\]

\[1)\ при\ x > 1:\]

\[(x + 1)^{2} - 4 \cdot (x - 1) \geq 13\]

\[x^{2} + 2x + 1 - 4x + 4 - 13 \geq 0\]

\[x^{2} - 2x - 8 \geq 0\]

\[D_{1} = 1 + 8 = 9\]

\[x_{1} = 1 + 3 = 4;\ \ \]

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

\[(x + 2)(x - 4) \geq 0\]

\[x \geq 4.\]

\[2)\ при\ x < 1:\]

\[(x + 1)^{2} + 4 \cdot (x - 1) \geq 13\]

\[x^{2} + 2x + 1 + 4x - 4 - 13 \geq 0\]

\[x^{2} + 6x - 16 \geq 0\]

\[D_{1} = 9 + 16 = 25\]

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

\[\ x_{2} = - 3 - 5 = - 8.\]

\[(x + 8)(x - 2) \geq 0\]

\[x \leq - 8.\]

\[Ответ:x \leq - 8;\ \ x \geq 4.\]