Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 28

\[\boxed{\mathbf{28.}}\]

\[\textbf{а)}\ 4^{\log_{4}(2x + 1)} = x^{2} + 3x - 5\]

\[2x + 1 > 0\]

\[2x > - 1\]

\[x > - 0,5.\]

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

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

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

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

\[x_{1} = - 3 < - 0,5;\ \ \]

\[x_{2} = 2.\]

\[Ответ:x = 2.\]

\[\textbf{б)}\ 5^{\log_{5}(x - 2)} = x^{2} + 4x - 30\]

\[x - 2 > 0\]

\[x > 2\]

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

\[x - 2 = x^{2} + 4x - 30\]

\[x^{2} + 3x - 28 = 0\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 28\]

\[x_{1} = - 7 < 2;\ \ \ \]

\[x_{2} = 4.\]

\[Ответ:x = 4.\]

\[\textbf{в)}\ 6^{\log_{6}(1 - x)} = x^{2} + 3x - 20\]

\[1 - x > 0\]

\[x < 1.\]

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

\[1 - x = x^{2} + 3x - 20\]

\[x^{2} + 4x - 21 = 0\]

\[D_{1} = 4 + 21 = 25\]

\[x_{1} = - 2 + 5 = 3 > 1;\]

\[x_{2} = - 2 - 5 = - 7.\]

\[Ответ:x = - 7.\]

\[\textbf{г)}\ 7^{\log_{7}(2 - x)} = x^{2} - 3x - 13\]

\[2 - x > 0\]

\[x < 2.\]

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

\[2 - x = x^{2} - 3x - 13\]

\[x^{2} - 2x - 15 = 0\]

\[D_{1} = 1 + 15 = 16\]

\[x_{1} = 1 + 4 = 5 > 2;\]

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

\[Ответ:x = - 3.\]