ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 158

\[\boxed{\mathbf{158.}}\]

\[1)\ \sqrt{5 - x} - \sqrt{5 + x} = 2\]

\[\left( \sqrt{5 - x} - \sqrt{5 + x} \right)^{2} = 4\]

\[6 = 2\sqrt{25 - x^{2}}\]

\[3 = \sqrt{25 - x^{2}}\]

\[9 = 25 - x^{2}\]

\[x^{2} = 16\]

\[x = \pm 4.\]

\[Проверим:\]

\[\sqrt{5 - ( - 4)} - \sqrt{5 - 4} =\]

\[= \sqrt{5 + 4} - \sqrt{1} = \sqrt{9} - 1 =\]

\[= 3 - 1 = 2;\]

\[\sqrt{5 - 4} - \sqrt{5 + 4} = \sqrt{1} - \sqrt{9} =\]

\[= 1 - 3 = - 2.\]

\[Ответ:\ \ x = - 4.\]

\[2)\ \sqrt{12 + x} - \sqrt{1 - x} = 1\]

\[\left( \sqrt{12 + x} - \sqrt{1 - x} \right)^{2} = 1\]

\[12 = 2\sqrt{12 - 12x + x - x^{2}}\]

\[6 = \sqrt{12 - 11x - x^{2}}\]

\[36 = 12 - 11x - x^{2}\]

\[x^{2} + 11x + 24 = 0\]

\[D = 11^{2} - 4 \bullet 24 =\]

\[= 121 - 96 = 25\]

\[x_{1} = \frac{- 11 - 5}{2} = - 8;\ \ \]

\[x_{2} = \frac{- 11 + 5}{2} = - 3.\]

\[Проверим:\]

\[\sqrt{12 - 8} - \sqrt{1 - ( - 8)} =\]

\[= \sqrt{4} - \sqrt{1 + 8} = 2 - \sqrt{9} =\]

\[= 2 - 3 = - 1;\]

\[\sqrt{12 - 3} - \sqrt{1 - ( - 3)} =\]

\[= \sqrt{9} - \sqrt{1 + 3} = \sqrt{3} - \sqrt{4} =\]

\[= 3 - 2 = 1.\]

\[Ответ:\ \ x = - 3.\]

\[3)\ \sqrt{x - 2} + \sqrt{x + 6} = 0\]

\[x - 2 = 0\]

\[x = 2.\]

\[x + 6 = 0\]

\[x = - 6.\]

\[Ответ:\ \ корней\ нет.\]

\[4)\ \sqrt{x + 7} + \sqrt{x - 2} = 9\]

\[\left( \sqrt{x + 7} + \sqrt{x - 2} \right)^{2} = 81\]

\[2\sqrt{x^{2} - 2x + 7x - 14} = 76 - 2x\]

\[\sqrt{x^{2} + 5x - 14} = 38 - x\]

\[x^{2} + 5x - 14 = (38 - x)^{2}\]

\[x^{2} + 5x - 14 = 1444 - 76x + x^{2}\]

\[81x = 1458\]

\[x = 18.\]

\[Проверим:\]

\[\sqrt{18 + 7} + \sqrt{18 - 2} =\]

\[= \sqrt{25} + \sqrt{16} = 5 + 4 = 9.\]

\[Ответ:\ \ x = 18.\]