Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 31

\[\boxed{\mathbf{31.}}\]

\[\textbf{а)}\ x - a + \sqrt{x - a} =\]

\[= 2x + 1 + \sqrt{x - a}\]

\[1)\ x - a \geq 0\]

\[x \geq a.\]

\[2)\ x - a - 2x - 1 = 0\]

\[- x - a - 1 = 0\]

\[x = - a - 1.\]

\[3) - a - 1 \geq a\]

\[2a \leq - 1\]

\[a \leq - \frac{1}{2}.\]

\[Ответ:при\ a \leq - \frac{1}{2}.\]

\[\textbf{б)}\ \frac{a - 2}{ax - 3} = 1\]

\[1)\ ax - 3 \neq 0\]

\[\text{ax} \neq 3\]

\[x \neq \frac{3}{a}.\]

\[2)\ a - 2 = ax - 3\]

\[ax = a - 2 + 3\]

\[ax = a + 1\]

\[x = \frac{a + 1}{a} = 1 + \frac{1}{a};\ \ a \neq 0.\]

\[3)\ 1 + \frac{1}{a} \neq \frac{3}{a}\]

\[\frac{2}{a} \neq 1\]

\[a \neq 2.\]

\[Ответ:при\ a \neq 0;a \neq 2.\]

\[\textbf{в)}\ \frac{1}{\left( \log_{x}5 \right)^{2}} - 2a \cdot \log_{5}x + 1 = 0\]

\[1)\ x > 0;\ \ x \neq 1;\]

\[2)\log_{5}x = t:\]

\[\frac{1}{\left( \frac{1}{t} \right)^{2}} - 2at + 1 =\]

\[= t^{2} - 2at + 1 = 0\]

\[D_{1} = a^{2} - 1\]

\[Уравнение\ имеет\ единственный\]

\[\ корень\ при\ D = 0.\]

\[a^{2} - 1 = 0\]

\[a^{2} = 1\]

\[a = \pm 1.\]

\[3)\ a = 1:\]

\[t = - \frac{b}{2a} = 1;\]

\[a = - 1:\]

\[t = - \frac{b}{2a} = - 1.\]

\[4)\ Проверка.\]

\[\log_{5}x = 1\]

\[x = 5.\]

\[\log_{5}x = - 1\]

\[x = \frac{1}{5}.\]

\[Ответ:\ при\ a = \pm 1.\]