Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 37

\[\boxed{\mathbf{37.}}\]

\[\textbf{а)}\lg(x - 1) + \sqrt{9 - x^{2}} =\]

\[= \lg{(x - 1)} + \sqrt{x + 5}\]

\[9 - x^{2} \geq 0\]

\[x^{2} - 9 \leq 0\]

\[(x + 3)(x - 3) \leq 0\]

\[- 3 \leq x \leq 3.\]

\[x + 5 \geq 0\]

\[x \geq - 5.\]

\[x - 1 > 0\]

\[x > 1.\]

\[M = (1;3\rbrack.\]

\[\sqrt{9 - x^{2}} = \sqrt{x + 5}\]

\[9 - x^{2} = x + 5\]

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

\[D = 1 + 16 = 17\]

\[x_{1} = \frac{- 1 + \sqrt{17}}{2};\]

\[x_{2} = \frac{- 1 - \sqrt{17}}{2} < 1.\]

\[Ответ:\ \frac{- 1 + \sqrt{17}}{2}.\]

\[\textbf{б)}\lg(1 - x) + \sqrt{25 - x^{2}} =\]

\[= \lg{(1 - x)} + \sqrt{x + 16}\]

\[25 - x^{2} \geq 0\]

\[x^{2} - 25 \leq 0\]

\[(x + 5)(x - 5) \leq 0\]

\[- 5 \leq x \leq 5.\]

\[x + 16 \geq 0\]

\[x \geq - 16.\]

\[1 - x > 0\]

\[x < 1.\]

\[M = \lbrack - 5;1).\]

\[\sqrt{25 - x^{2}} = \sqrt{x + 16}\]

\[25 - x^{2} = x + 16\]

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

\[D = 1 + 36 = 37\]

\[x_{1} = \frac{- 1 - \sqrt{37}}{2};\]

\[x_{2} = \frac{- 1 + \sqrt{37}}{2} > 1.\]

\[Ответ:\ x = \frac{- 1 - \sqrt{37}}{2}.\]