Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 27

\[\boxed{\mathbf{27.}}\]

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

\[= 15 + \log_{2}{(x + 1)}\]

\[x + 1 > 0\]

\[x > - 1.\]

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

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

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

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

\[x_{2} = - 1 - 4 = - 5 < - 1.\]

\[Ответ:x = 3.\]

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

\[= 7 - \log_{3}(1 - x)\]

\[1 - x > 0\]

\[x < 1.\]

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

\[x^{2} - 6x - 7 = 0\]

\[D_{1} = 9 + 7 = 16\]

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

\[x_{2} = 3 - 4 = - 1.\]

\[Ответ:x = - 1.\]

\[\textbf{в)}\ x^{2} + \log_{4}x = 7x + \log_{4}x\]

\[x > 0\]

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

\[x^{2} = 7x\]

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

\[x(x - 7) = 0\]

\[x = 0 - не\ подходит;\]

\[x - 7 = 0\]

\[x = 7.\]

\[Ответ:x = 7.\]

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

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

\[- x > 0\]

\[x < 0\]

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

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

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

\[x(x + 6) = 0\]

\[x = 0 - не\ подходит;\]

\[x + 6 = 0\]

\[x = - 6.\]

\[Ответ:x = - 6.\]