Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 33

\[\boxed{\mathbf{33.}}\]

\[\textbf{а)}\ \sqrt{x + 2} + \sqrt{2x - 3} = \sqrt{3x + 3}\]

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

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

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

\[(x + 2)(2x - 3) = 4\]

\[2x^{2} + 4x - 3x - 6 - 4 = 0\]

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

\[D = 1 + 80 = 81\]

\[x_{1} = \frac{- 1 + 9}{4} = 2;\]

\[x_{2} = \frac{- 1 - 9}{4} = - 2,5.\]

\[Проверка.\]

\[x = 2:\]

\[\sqrt{2 + 2} + \sqrt{4 - 3} = \sqrt{6 + 3}\]

\[2 + 1 = 3\]

\[3 = 3.\]

\[x = - 2,5:\]

\[\sqrt{- 2,5 + 2} < 0\]

\[не\ корень.\]

\[Ответ:x = 2.\]

\[\textbf{б)}\ \sqrt{x + 1} + \sqrt{x + 6} =\]

\[= \sqrt{2x + 19}\]

\[x + 1 + 2\sqrt{x + 1}\sqrt{x + 6} + x + 6 = 2x + 19\]

\[2\sqrt{x + 1}\sqrt{x + 6} = 12\]

\[\sqrt{x + 1}\sqrt{x + 6} = 6\]

\[(x + 1)(x + 6) = 36\]

\[x^{2} + x + 6x + 6 - 36 = 0\]

\[x^{2} + 7x - 30 = 0\]

\[x_{1} + x_{2} = - 7;\ \ x_{1} \cdot x_{2} = - 30\]

\[x_{1} = - 10;\ \ x_{2} = 3.\]

\[Проверка.\]

\[x = 3:\]

\[\sqrt{3 + 1} + \sqrt{3 + 6} = \sqrt{6 + 19}\]

\[2 + 3 = 5\]

\[5 = 5.\]

\[x = - 10:\]

\[\sqrt{- 10 + 1} < 0\]

\[нет\ решений.\]

\[Ответ:x = 3.\]

\[\textbf{в)}\ \sqrt{6x + 1} - \sqrt{x - 3} = \sqrt{3x + 4}\]

\[6x + 1 - 2\sqrt{6x + 1}\sqrt{x - 3} + x - 3 = 3x + 4\]

\[2\sqrt{6x + 1}\sqrt{x - 3} = 4x - 6\]

\[\sqrt{6x + 1}\sqrt{x - 3} = 2x - 3\]

\[(6x + 1)(x - 3) = (2x - 3)^{2}\]

\[6x^{2} + x - 18x - 3 =\]

\[= 4x^{2} - 12x + 9\]

\[2x^{2} - 5x - 12 = 0\]

\[D = 25 + 96 = 121\]

\[x_{1} = \frac{5 + 11}{4} = 4;\]

\[x_{2} = \frac{5 - 11}{4} = - 1,5.\]

\[Проверка.\]

\[x = 4:\]

\[\sqrt{24 + 1} - \sqrt{4 - 3} = \sqrt{12 + 4}\]

\[5 - 1 = 4\]

\[4 = 4.\]

\[x = - 1,5:\]

\[\sqrt{- 9 + 1} < 0\]

\[не\ корень.\]

\[Ответ:x = 4.\]

\[\textbf{г)}\ \sqrt{9 - 5x} - \sqrt{x - 1} = 2\sqrt{2 - x}\]

\[- 2\sqrt{9 - 5x}\sqrt{x - 1} = 0\]

\[\sqrt{9 - 5x}\sqrt{x - 1} = 0\]

\[(9 - 5x)(x - 1) = 0\]

\[x = 1;\ \ x = \frac{9}{5} = 1,8.\]

\[Проверка.\]

\[x = 1:\]

\[\sqrt{9 - 5} - \sqrt{1 - 1} = 2\sqrt{2 - 1}\]

\[2 - 0 = 2\]

\[2 = 2.\]

\[x = 1,8:\]

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

\[\sqrt{0} - \sqrt{0,8} = 2 \cdot \sqrt{0,2}\]

\[не\ корень.\]

\[Ответ:x = 1.\]