ГДЗ по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 24

\[\boxed{\mathbf{24}\mathbf{.}}\]

\[\textbf{а)}\lg\left( x^{2} - 4 \right) = \lg{(x - 1)}\]

\[x^{2} - 4 > 0\]

\[(x + 2)(x - 2) > 0\]

\[x < - 2;\ \ x > 2.\]

\[x - 1 > 0\]

\[x > 1.\]

\[M = (2; + \infty).\]

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

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

\[D = 1 + 12 = 13\]

\[x_{1} = \frac{1 + \sqrt{13}}{2} > 2;\]

\[x_{2} = \frac{1 - \sqrt{13}}{2} < 2.\]

\[Ответ:x = \frac{1 + \sqrt{13}}{2}.\]

\[\textbf{б)}\log_{2}\left( x^{2} - 9 \right) =\]

\[= \log_{2}(2 - x) + 1\]

\[\log_{2}\left( x^{2} - 9 \right) =\]

\[= \log_{2}{(2 - x)} + \log_{2}2^{1}\]

\[\log_{2}\left( x^{2} - 9 \right) = \log_{2}{2(2 - x)}\]

\[x^{2} - 9 > 0\]

\[(x + 3)(x - 3) > 0\]

\[x < - 3;\ \ x > 3.\]

\[2 - x > 0\]

\[x < 2.\]

\[M = ( - \infty;\ - 3).\]

\[x^{2} - 9 = 2(2 - x)\]

\[x^{2} - 9 = 4 - 2x\]

\[x^{2} + 2x - 13 = 0\]

\[D_{1} = 1 + 13 = 14\]

\[x_{1} = - 1 + \sqrt{14} > - 3;\]

\[x_{2} = - 1 - \sqrt{14} < - 3.\]

\[Ответ:x = - 1 - \sqrt{14}.\]

\[\textbf{в)}\lg{3x^{2}} = \lg(2x + 1)\]

\[3x^{2} \neq 0\]

\[x \neq 0.\]

\[2x + 1 > 0\]

\[2x > - 1\]

\[x > - 0,5.\]

\[M = ( - 0,5;0) \cup (0; + \infty).\]

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

\[3x^{2} - 2x - 1 = 0\]

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

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

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

\[Ответ:x = - \frac{1}{3};\ \ x = 1.\]

\[\textbf{г)}\log_{2}\left( 16 - x^{2} \right) =\]

\[= \log_{2}{(1 + x)} + 1\]

\[\log_{2}\left( 16 - x^{2} \right) =\]

\[= \log_{2}{(1 + x)} + \log_{2}2^{1}\]

\[\log_{2}{(16 - x^{2})} = \log_{2}{2(1 + x)}\]

\[16 - x^{2} > 0\]

\[x^{2} - 16 < 0\]

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

\[- 4 < x < 4.\]

\[1 + x > 0\]

\[x > - 1.\]

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

\[16 - x^{2} = 2(1 + x)\]

\[16 - x^{2} = 2 + 2x\]

\[x^{2} + 2x - 14 = 0\]

\[D_{1} = 1 + 14 = 15\]

\[x_{1} = - 1 + \sqrt{15};\]

\[x_{2} = - 1 - \sqrt{15} < - 1.\]

\[Ответ:x = - 1 + \sqrt{15}.\]