Решебник по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 32

\[\boxed{\mathbf{32.}}\]

\[\textbf{а)}\ \sqrt{27} \cdot 3^{- 6x^{2}} \geq 9^{4x}\]

\[\left( 3^{3} \right)^{\frac{1}{2}} \cdot 3^{- 6x^{2}} \geq \left( 3^{2} \right)^{4x}\]

\[3^{\frac{3}{2} - 6x^{2}} \geq 3^{8x}\]

\[\frac{3}{2} - 6x^{2} \geq 8x\]

\[6x^{2} + 8x - \frac{3}{2} \leq 0\ \ \ | \cdot 2\]

\[12x^{2} + 16x - 3 \leq 0\]

\[D_{1} = 64 + 36 = 100\]

\[x_{1} = \frac{- 8 + 10}{12} = \frac{2}{12} = \frac{1}{6};\]

\[x_{2} = \frac{- 8 - 10}{12} = - \frac{18}{12} = - \frac{3}{2};\]

\[\left( x + \frac{3}{2} \right)\left( x - \frac{1}{6} \right) \leq 0\]

\[- \frac{3}{2} \leq x \leq \frac{1}{6}.\]

\[\textbf{б)}\ \sqrt{32} \cdot 2^{- 4x^{2}} \geq 8^{3x}\]

\[\left( 2^{5} \right)^{\frac{1}{2}} \cdot 2^{- 4x^{2}} \geq \left( 2^{3} \right)^{3x}\]

\[2^{\frac{5}{2}} \cdot 2^{- 4x^{2}} \geq 2^{9x}\]

\[2^{\frac{5}{2} - 4x^{2}} \geq 2^{9x}\]

\[\frac{5}{2} - 4x^{2} \geq 9x\]

\[4x^{2} + 9x - \frac{5}{2} \leq 0\ \ \ | \cdot 2\]

\[8x^{2} + 18x - 5 \leq 0\]

\[D_{1} = 81 + 40 = 121\]

\[x_{1} = \frac{- 9 - 11}{8} = - \frac{20}{8} =\]

\[= - \frac{5}{4} = - 1,25;\]

\[x_{2} = \frac{- 9 + 11}{8} = \frac{2}{8} = \frac{1}{4} = 0,25;\ \]

\[(x + 1,25)(x - 0,25) \leq 0\]

\[- 1,25 \leq x \leq 0,25.\]

\[\textbf{в)}\ 4 \cdot \left( \frac{1}{2} \right)^{5x^{2}} \leq \left( \frac{1}{8} \right)^{- 3x}\]

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

\[2^{2} \cdot 2^{- 5x^{2}} \leq \left( 2^{- 3} \right)^{- 3x}\]

\[2^{2 - 5x^{2}} \leq 2^{9x}\]

\[2 - 5x^{2} \leq 9x\]

\[5x^{2} + 9x - 2 \geq 0\]

\[D = 81 + 40 = 121\]

\[x_{1} = \frac{- 9 + 11}{10} = \frac{2}{10} = \frac{1}{5} = 0,2;\]

\[x_{2} = \frac{- 9 - 11}{10} = - \frac{20}{10} = - 2;\]

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

\[x \leq - 2;\ \ x \geq 0,5.\]

\[\textbf{г)}\ 125 \cdot \left( \frac{1}{5} \right)^{x^{2}} \leq \left( \frac{1}{25} \right)^{- 4x}\]

\[5^{3} \cdot \left( 5^{- 1} \right)^{x^{2}} \leq \left( 5^{- 2} \right)^{- 4x}\]

\[5^{3} \cdot 5^{- x^{2}} \leq 5^{8x}\]

\[5^{3 - x^{2}} \leq 5^{8x}\]

\[3 - x^{2} \leq 8x\]

\[x^{2} + 8x - 3 \geq 0\]

\[D_{1} = 16 + 3 = 19\]

\[x_{1} = - 4 + \sqrt{19};\]

\[x_{2} = - 4 - \sqrt{19}.\]

\[\left( x + 4 + \sqrt{19} \right)\left( x + 4 - \sqrt{19} \right) \geq 0\]

\[x \leq - 4 - \sqrt{19};\]

\[x \geq - 4 + \sqrt{19}.\]