ГДЗ по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 35

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

\[x - 5 \neq 0\]

\[x \neq 5.\]

\[x > 0.\]

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

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

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

\[x^{2} - 10x + 25 - x = 0\]

\[x^{2} - 11x + 25 = 0\]

\[D = 121 - 100 = 21\]

\[x_{1} = \frac{11 + \sqrt{21}}{2} > 5;\]

\[x_{2} = \frac{11 - \sqrt{21}}{2} > 0 < 5.\]

\[Ответ:x = \frac{11 \pm \sqrt{21}}{2}.\]

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

\[x - 3 \neq 0\]

\[x \neq 3.\]

\[x > 0.\]

\[M = (0;3) \cup (3; + \infty).\]

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

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

\[x^{2} - 6x + 9 = x\]

\[x^{2} - 7x + 9 = 0\]

\[D = 49 - 36 = 13\]

\[x_{1} = \frac{7 + \sqrt{13}}{2} > 3;\]

\[x_{2} = \frac{7 - \sqrt{13}}{2} > 0 < 3.\]

\[Ответ:x = \frac{7 \pm \sqrt{13}}{2}.\]

\[\textbf{в)}\lg(x - 2)^{2} = 2\lg\sqrt{3 - x}\]

\[x - 2 \neq 0\]

\[x \neq 2.\]

\[3 - x > 0\]

\[x < 3.\]

\[M = ( - \infty;2) \cup (2;3).\]

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

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

\[x^{2} - 4x + 4 - 3 + x = 0\]

\[x^{2} - 3x + 1 = 0\]

\[D = 9 - 4 = 5\]

\[x_{1} = \frac{3 + \sqrt{5}}{2} > 2 < 3;\]

\[x_{2} = \frac{3 - \sqrt{5}}{2} > 0 < 2.\]

\[Ответ:x = \frac{3 \pm \sqrt{5}}{2}.\]

\[\textbf{г)}\lg(x - 1)^{2} = 2\lg\sqrt{2 - x}\]

\[x - 1 \neq 0\]

\[x \neq 1.\]

\[2 - x > 0\]

\[x < 2.\]

\[M = ( - \infty;1) \cup (1;2).\]

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

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

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

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

\[D = 1 + 4 = 5\]

\[x_{1} = \frac{1 + \sqrt{5}}{2} > 1 < 2;\]

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

\[Ответ:x = \frac{1 \pm \sqrt{5}}{2}.\]