ГДЗ по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 10

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Год:2020-2021-2022
Тип:учебник

Задание 10

\[\boxed{\mathbf{10.}}\]

\[\textbf{а)}\ \frac{4^{1 - 2x}}{8} = 0,5 \cdot 2^{1,2 + 2x}\]

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

\[2^{2 - 4x - 3} = 2^{- 1 + 1,2 + 2x}\]

\[2^{- 4x - 1} = 2^{0,2 + 2x}\]

\[- 4x - 1 = 0,2 + 2x\]

\[- 6x = 1,2\]

\[x = - 0,2.\]

\[\textbf{б)}\ 0,2 \cdot 5^{0,2x + 3} = \frac{25^{0,2 - x}}{125}\]

\[5^{- 1} \cdot 5^{0,2x + 3} = \frac{\left( 5^{2} \right)^{0,2 - x}}{5^{3}}\]

\[5^{- 1 + 0,2x + 3} = 5^{0,4 - 2x - 3}\]

\[5^{0,2x + 2} = 5^{- 2x - 2,6}\]

\[0,2x + 2 = - 2x - 2,6\]

\[2,2x = - 4,6\]

\[x = - \frac{46}{22} = - \frac{23}{11}\]

\[x = - 2\frac{1}{11}.\]

\[\textbf{в)}\ \frac{3^{1 + 3x}}{9} = 27 \cdot 3^{1 - 2x}\]

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

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

\[3^{3x - 1} = 3^{4 - 2x}\]

\[3x - 1 = 4 - 2x\]

\[5x = 5\]

\[x = 1.\]

\[\textbf{г)}\ \frac{2}{3} \cdot 4^{x - 2} = \frac{8^{3 - 2x}}{12}\]

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

\[\frac{2^{1 + 2x - 4}}{3} = \frac{2^{9 - 6x - 2}}{3}\]

\[2^{2x - 3} = 2^{7 - 6x}\]

\[2x - 3 = 7 - 6x\]

\[8x = 10\]

\[x = \frac{10}{8} = \frac{5}{4}\]

\[x = 1,25.\]

\[\textbf{д)}\ (0,81)^{- 2x} = \left( \frac{\sqrt{10}}{3} \right)^{3x^{2} - 3}\]

\[\left( \frac{81}{100} \right)^{- 2x} = \left( \frac{\sqrt{10}}{3} \right)^{3x^{2} - 3}\]

\[\left( \left( \frac{3}{\sqrt{10}} \right)^{4} \right)^{- 2x} = \left( \frac{\sqrt{10}}{3} \right)^{3x^{2} - 3}\]

\[\left( \left( \frac{\sqrt{10}}{3} \right)^{- 4} \right)^{- 2x} = \left( \frac{\sqrt{10}}{3} \right)^{3x^{2} - 3}\]

\[\left( \frac{\sqrt{10}}{3} \right)^{8x} = \left( \frac{\sqrt{10}}{3} \right)^{3x^{2} - 3}\]

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

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

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

\[x_{1} = \frac{4 + 5}{3} = 3;\]

\[x_{2} = \frac{4 - 5}{3} = - \frac{1}{3}.\]

\[\textbf{е)}\ \left( \frac{2}{9} \right)^{x^{2} - 1} = \left( \frac{3}{\sqrt{2}} \right)^{1,5x}\]

\[\left( \left( \frac{\sqrt{2}}{3} \right)^{2} \right)^{x^{2} - 1} = \left( \frac{3}{\sqrt{2}} \right)^{1,5x}\]

\[\left( \left( \frac{3}{\sqrt{2}} \right)^{- 2} \right)^{x^{2} - 1} = \left( \frac{3}{\sqrt{2}} \right)^{1,5x}\]

\[\left( \frac{3}{\sqrt{2}} \right)^{- 2x^{2} + 2} = \left( \frac{3}{\sqrt{2}} \right)^{1,5x}\]

\[- 2x^{2} + 2 = 1,5x\]

\[2x^{2} + 1,5x - 2 = 0\ \ \ | \cdot 2\]

\[4x^{2} + 3x - 4 = 0\]

\[D = 9 + 64 = 73\]

\[x = \frac{- 3 \pm \sqrt{73}}{8}.\]

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