Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 707

\[\boxed{\mathbf{707}.}\]

\[1)\ 3^{x + 2} + 3^{x - 1} < 28;\]

\[3^{x} \bullet \left( 3^{2} + 3^{- 1} \right) < 28;\]

\[3^{x} \bullet \left( 9 + \frac{1}{3} \right) < 28;\]

\[3^{x} \bullet \left( \frac{27}{3} + \frac{1}{3} \right) < 28;\]

\[3^{x} \bullet \frac{28}{3} < 28;\]

\[3^{x} < 3;\]

\[3^{x} < 3^{1};\ \]

\[x < 1;\]

\[Ответ:\ \ x < 1.\]

\[2)\ 2^{x - 1} + 2^{x + 3} > 17;\]

\[2^{x} \bullet \left( 2^{- 1} + 2^{3} \right) > 17;\]

\[2^{x} \bullet \left( \frac{1}{2} + 8 \right) > 17;\]

\[2^{x} \bullet \left( \frac{1}{2} + \frac{16}{2} \right) > 17;\]

\[2^{x} \bullet \frac{17}{2} > 17;\]

\[2^{x} > 2;\]

\[2^{x} > 2^{1};\]

\[x > 1;\]

\[Ответ:\ \ x > 1.\]

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

\[+ 2^{2x - 3} \geq 448;\]

\[2^{2x} \bullet \left( 2^{- 1} + 2^{- 2} + 2^{- 3} \right) \geq 448;\]

\[2^{2x} \bullet \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \right) \geq 448;\]

\[2^{2x} \bullet \left( \frac{4}{8} + \frac{2}{8} + \frac{1}{8} \right) \geq 448;\]

\[2^{2x} \bullet \frac{7}{8} \geq 448;\]

\[2^{2x} \geq 512;\]

\[2^{2x} \geq 2^{9};\]

\[2x \geq 9;\]

\[x \geq 4,5;\]

\[Ответ:\ \ x \geq 4,5.\]

\[4)\ 5^{3x + 1} - 5^{3x - 3} \leq 624;\]

\[5^{3x} \bullet \left( 5^{1} - 5^{- 3} \right) \leq 624;\]

\[5^{3x} \bullet \left( 5 - \frac{1}{125} \right) \leq 624;\]

\[5^{3x} \bullet \left( \frac{625}{125} - \frac{1}{125} \right) \leq 624;\]

\[5^{3x} \bullet \frac{624}{125} \leq 624;\]

\[5^{3x} \leq 125;\]

\[5^{3x} \leq 5^{3};\]

\[3x \leq 3;\]

\[x \leq 1;\]

\[Ответ:\ \ x \leq 1.\]