Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 44

\[\boxed{\mathbf{44.}}\]

\[\textbf{а)}\ \frac{1}{\sqrt{x - 1}} = (x - 1)^{\log_{\frac{1}{25}}\left( 8 + 2x - x^{2} \right)}\]

\[x - 1 > 0\]

\[x > 1.\]

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

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

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

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

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

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

\[- 2 < x < 4.\]

\[M = (1;4).\]

\[(x - 1)^{- \frac{1}{2}} = (x - 1)^{\log_{\frac{1}{25}}\left( 8 + 2x - x^{2} \right)}\]

\[1)\lg(x - 1) = 0\]

\[x - 1 = 1\]

\[x = 2.\]

\[2)\ \log_{\frac{1}{25}}\left( 8 + 2x - x^{2} \right) = - \frac{1}{2}\]

\[\left( \frac{1}{25} \right)^{- \frac{1}{2}} = 8 + 2x - x^{2}\]

\[\left( 5^{- 2} \right)^{- \frac{1}{2}} = 8 + 2x - x^{2}\]

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

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

\[D_{1} = 1 + 3 = 4\]

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

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

\[Ответ:x = 2;x = 3.\]

\[\textbf{б)}\ \frac{1}{\sqrt{2x - 1}} =\]

\[= (2x - 1)^{\log_{\frac{1}{4}}\left( 1 + 7x - 2x^{2} \right)}\]

\[2x - 1 > 0\]

\[x > 0,5.\]

\[1 + 7x - 2x^{2} > 0\]

\[2x^{2} - 7x - 1 < 0\]

\[D = 49 + 8 = 57\]

\[x_{1,2} = \frac{7 \pm \sqrt{57}}{4};\]

\[\frac{7 - \sqrt{57}}{4} < x < \frac{7 + \sqrt{57}}{4}.\]

\[M = \left( \frac{1}{2};\ \frac{7 + \sqrt{57}}{4} \right).\]

\[1)\ \lg(2x - 1) = 0\]

\[2x - 1 = 1\]

\[2x = 2\]

\[x = 1.\]

\[2)\ \log_{\frac{1}{4}}\left( 1 + 7x - 2x^{2} \right) = - \frac{1}{2}\]

\[\left( \frac{1}{4} \right)^{- \frac{1}{2}} = 1 + 7x - 2x^{2}\]

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

\[2 = 1 + 7x - 2x^{2}\]

\[2x^{2} - 7x + 1 = 0\]

\[D = 49 - 8 = 41\]

\[x_{1,2} = \frac{7 \pm \sqrt{41}}{4}.\]

\[Ответ:x = 1;x = \frac{7 + \sqrt{41}}{4}.\]