Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 36

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\[\textbf{а)}\ 2\log_{4}{(x + 1)} = \log_{4}{(4x + 9)}\]

\[\log_{4}\left( (x + 1)^{2} \right) = \log_{4}{(4x + 9)}\]

\[x + 1 > 0\]

\[x > - 1.\]

\[(x + 1)^{2} = 4x + 9\]

\[x^{2} + 2x + 1 - 4x - 9 = 0\]

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

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

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

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

\[= - 2 < - 1\ (не\ корень).\]

\[Ответ:x = 4.\]

\[\textbf{б)}\ 2\log_{5}{(x - 1)} = \log_{5}{(7 - x)}\]

\[\log_{5}(x - 1)^{2} = \log_{5}(7 - x)\]

\[x - 1 > 0\]

\[x > 1.\]

\[7 - x > 0\]

\[x < 7.\]

\[1 < x < 7.\]

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

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

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

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

\[x_{1} = 3;\]

\[x_{2} = - 2\ (не\ подходит).\]

\[Ответ:x = 3.\]

\[\textbf{в)}\log_{7}(x - 2) + \log_{7}(x + 3) =\]

\[= \log_{7}{(2x^{2} - 4x)}\]

\[\log_{7}\left( (x - 2)(x + 3) \right) =\]

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

\[x > 2.\]

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

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

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

\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = 6\]

\[x_{1} = 3;\]

\[x_{2} = 2\ (не\ подходит).\]

\[Ответ:x = 3.\]

\[\textbf{г)}\log_{6}(x + 2) + \log_{6}(x - 3) =\]

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

\[\log_{6}\left( (x + 2)(x - 3) \right) =\]

\[= \log_{6}\left( 2x^{2} - 5x - 6 \right)\]

\[x > 3.\]

\[(x + 2)(x - 3) = 2x^{2} - 5x - 6\]

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

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

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

\[x(x - 4) = 0\]

\[x = 0 < 3 - не\ подходит.\]

\[x = 4.\]

\[Ответ:x = 4.\]

\[\textbf{д)}\lg(x - 2) - \lg(x + 3) =\]

\[= \lg\frac{x}{5x + 3}\]

\[x > 2;\]

\[\lg\left( \frac{x - 2}{x + 3} \right) = \lg\frac{x}{5x + 3}\]

\[\frac{x - 2}{x + 3} = \frac{x}{5x + 3}\]

\[x \neq - 3;\ \ x \neq - \frac{3}{5};\ \]

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

\[5x^{2} - 10x + 3x - 6 = x^{2} + 3x\]

\[4x^{2} - 10x - 6 = 0\ \ \ |\ :2\]

\[2x^{2} - 5x - 3 = 0\]

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{5 + 7}{4} = 3;\]

\[x_{2} = \frac{5 - 7}{4} =\]

\[= - \frac{1}{2} < 2\ (не\ корень).\]

\[Ответ:x = 3.\]

\[\textbf{е)}\lg(x + 2) - \lg(x - 3) =\]

\[= \lg\frac{5x + 4}{x}\]

\[x > 3;\ \ x \neq 0.\]

\[\lg\left( \frac{x + 2}{x - 3} \right) = \lg\frac{5x + 4}{x}\]

\[\frac{x + 2}{x - 3} = \frac{5x + 4}{x}\]

\[x(x + 2) = (x - 3)(5x + 4)\]

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

\[= 5x^{2} - 15x + 4x - 12\]

\[4x^{2} - 13x - 12 = 0\]

\[D = 169 + 192 = 361\]

\[x_{1} = \frac{13 + 19}{8} = 4;\]

\[x_{2} = \frac{13 - 19}{8} =\]

\[= - \frac{3}{4} < 3\ (не\ корень).\]

\[Ответ:x = 4.\]