ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 611

\[\boxed{\mathbf{611}.}\]

\[1)\ \sqrt{1 - 2x} - \sqrt{13 + x} =\]

\[= \sqrt{x + 4};\]

\[1 - 2x - 2\sqrt{(1 - 2x)(13 + x)} +\]

\[+ 13 + x = x + 4;\]

\[10 - 2x =\]

\[= 2\sqrt{13 + x - 26x - 2x^{2}};\]

\[5 - x = \sqrt{13 - 25x - 2x^{2}};\]

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

\[= 13 - 25x - 2x^{2};\]

\[3x^{2} + 15x + 12 = 0;\]

\[x^{2} + 5x + 4 = 0;\]

\[D = 5^{2} - 4 \bullet 4 = 25 - 16 = 9\]

\[x_{1} = \frac{- 5 - 3}{2} = - 4\ \ и\ \ \]

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

\[Выполним\ проверку:\]

\[\sqrt{1 + 2 \bullet 4} - \sqrt{13 - 4} -\]

\[- \sqrt{- 4 + 4} = \sqrt{9} - \sqrt{9} - \sqrt{0} = 0;\]

\[\sqrt{1 - 2 \bullet 1} - \sqrt{13 + 1} - \sqrt{1 + 4} =\]

\[= \sqrt{- 1}\ldots - не\ имеет\ смысла.\]

\[Ответ:\ \ x = - 4.\]

\[2)\ \sqrt{7x + 1} - \sqrt{6 - x} =\]

\[= \sqrt{15 + 2x};\]

\[7x + 1 - 2\sqrt{(7x + 1)(6 - x)} +\]

\[+ 6 - x = 15 + 2x;\]

\[4x - 8 = 2\sqrt{42x - 7x^{2} + 6 - x};\]

\[2x - 4 = \sqrt{41x - 7x^{2} + 6};\]

\[4x^{2} - 16x + 16 = 41x -\]

\[- 7x^{2} + 6;\]

\[11x^{2} - 57x + 10 = 0;\]

\[D = 57^{2} - 4 \bullet 11 \bullet 10 =\]

\[= 3249 - 440 = 2809\]

\[x_{1} = \frac{57 - 53}{2 \bullet 11} = \frac{4}{2 \bullet 11} = \frac{2}{11};\]

\[x_{2} = \frac{57 + 53}{2 \bullet 11} = \frac{110}{2 \bullet 11} = \frac{10}{2} = 5;\]

\[Выполним\ проверку:\]

\[\sqrt{7 \bullet \frac{2}{11} + 1} - \sqrt{6 - \frac{2}{11}} -\]

\[- \sqrt{15 + 2 \bullet \frac{2}{11}} = \sqrt{\frac{25}{11}} - \sqrt{\frac{64}{11}} -\]

\[- \sqrt{\frac{169}{11}} = - \frac{16}{\sqrt{11}};\]

\[\sqrt{7 \bullet 5 + 1} - \sqrt{6 - 5} -\]

\[- \sqrt{15 + 2 \bullet 5} = \sqrt{36} - \sqrt{1} -\]

\[- \sqrt{25} = 6 - 1 - 5 = 0.\]

\[Ответ:\ \ x = 5.\]