Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 34

\[\boxed{\mathbf{34.}}\]

\[\textbf{а)}\log_{2}(x + 2) + \log_{2}(3x + 2) =\]

\[= \log_{2}{(5x + 22)}\]

\[x + 2 > 0\]

\[x > - 2.\]

\[3x + 2 > 0\]

\[x > - \frac{2}{3}.\]

\[5x + 22 > 0\]

\[x > - 4,4.\]

\[M = \left( - \frac{2}{3}; + \infty \right).\]

\[\log_{2}{(x + 2)(3x + 2)} =\]

\[= \log_{2}{(5x + 22)}\]

\[(x + 2)(3x + 2) = 5x + 22\]

\[3x^{2} + 6x + 2x + 4 - 5x - 22 =\]

\[= 0\]

\[3x^{2} + 3x - 18 = 0\ \ \ |\ :3\]

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

\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 6\]

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

\[x_{2} = 2.\]

\[Ответ:x = 2.\]

\[\textbf{б)}\log_{3}(x - 5) + \log_{3}(x + 1) =\]

\[= \log_{3}{(3x + 3)}\]

\[x - 5 > 0\]

\[x > 5.\]

\[x + 1 > 0\]

\[x > - 1.\]

\[3x + 3 > 0\]

\[x > - 1.\]

\[M = (5; + \infty).\]

\[\log_{3}{(x - 5)(x + 1)} =\]

\[= \log_{3}(3x + 3)\]

\[(x - 5)(x + 1) = 3x + 3\]

\[(x - 5)(x + 1) = 3(x + 1)\]

\[x - 5 = 3\]

\[x = 8.\]

\[Ответ:x = 8.\]

\[\textbf{в)}\log_{5}(x - 6) + \log_{5}(2x + 11) =\]

\[= \log_{5}{(3x + 4)}\]

\[x - 6 > 0\]

\[x > 6.\]

\[2x + 11 > 0\]

\[x > - 5,5.\]

\[3x + 4 > 0\]

\[x > - \frac{4}{3}.\]

\[M = (6; + \infty).\]

\[\log_{5}{(x - 6)(2x + 11)} =\]

\[= \log_{5}{(3x + 4)}\]

\[(x - 6)(2x + 11) = 3x + 4\]

\[2x^{2} - 4x - 70 = 0\ \ \ |\ :2\]

\[x^{2} - 2x - 35 = 0\]

\[D_{1} = 1 + 35 = 36\]

\[x_{1} = 1 + 6 = 7;\]

\[x_{2} = 1 - 6 = - 5 < 6.\]

\[Ответ:x = 7.\]

\[2x + 6 > 0\]

\[x > - 3.\]

\[3x - 14 > 0\]

\[x > \frac{14}{3}\]

\[x > 4\frac{2}{3}.\]

\[3x + 1 > 0\]

\[x > - \frac{1}{3}.\]

\[M = \left( 4\frac{2}{3}; + \infty \right).\]

\[\log_{4}{(2x + 6)(3x - 14)} =\]

\[= \log_{4}{(3x + 1)}\]

\[(2x + 6)(3x - 14) = 3x + 1\]

\[6x^{2} - 13x - 85 = 0\]

\[D = 169 + 2040 = 2209 = 47^{2}\]

\[x_{1} = \frac{13 + 47}{12} = 5 > 4\frac{2}{3};\]

\[x_{2} = \frac{13 - 47}{12} = - \frac{17}{6} < 4\frac{2}{3}.\]

\[Ответ:x = 5.\]