Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 37

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

\[\textbf{а)}\ \sqrt{x - 5}\sqrt{x - 6} = \sqrt{x - 2}\]

\[(x - 5)(x - 6) = x - 2\]

\[x^{2} - 5x - 6x + 30 - x + 2 = 0\]

\[x^{2} - 12x + 32 = 0\]

\[D_{1} = 36 - 32 = 4\]

\[x_{1} = 6 + 2 = 8;\]

\[x_{2} = 6 - 2 = 4.\]

\[Проверка.\]

\[x = 8:\]

\[\sqrt{8 - 5}\sqrt{8 - 6} = \sqrt{8 - 2}\]

\[\sqrt{3} \cdot \sqrt{2} = \sqrt{6}.\]

\[x = 4:\]

\[\sqrt{4 - 5} < 0 - не\ корень.\]

\[Ответ:x = 8.\]

\[\textbf{б)}\ \sqrt{x - 3}\sqrt{x - 7} = \sqrt{x + 3}\]

\[x > 7;\]

\[(x - 3)(x - 7) = x + 3\]

\[x^{2} - 3x - 7x + 21 - x - 3 = 0\]

\[x^{2} - 11x + 18 = 0\]

\[x_{1} + x_{2} = 11;\ \ x_{1} \cdot x_{2} = 18\]

\[x_{1} = 2 < 7 - не\ корень.\]

\[x_{2} = 9.\]

\[Ответ:x = 9.\]

\[\textbf{в)}\ \frac{\sqrt{7x - 21}}{\sqrt{x + 2}} = \frac{\sqrt{5x - 17}}{2}\]

\[\frac{7x - 21}{x + 2} = \frac{5x - 17}{4}\]

\[x \neq - 2;\ \ x > 3,4.\]

\[4(7x - 21) = (x + 2)(5x - 17)\]

\[28x - 84 =\]

\[= 5x^{2} + 10x - 17x - 34\]

\[5x^{2} - 35x + 50 = 0\ \ \ \ |\ :5\]

\[x^{2} - 7x + 10 = 0\]

\[x_{1} + x_{2} = 7;\ \ x_{1} \cdot x_{2} = 10\]

\[x_{1} = 2 < 3,4 - не\ корень;\ \ \ \]

\[x_{2} = 5.\]

\[Ответ:x = 5.\]

\[\textbf{г)}\ \frac{\sqrt{3x - 7}}{\sqrt{x + 3}} = \sqrt{3 - 4x}\]

\[x > \frac{7}{3} > 2\frac{1}{3};\]

\[x < \frac{3}{4}.\]

\[\frac{3x - 7}{x + 3} = 3 - 4x;\ \ x \neq - 3\]

\[3x - 7 = (x + 3)(3 - 4x)\]

\[3x - 7 = 3x + 9 - 4x^{2} - 12x\]

\[4x^{2} + 12x - 16 = 0\ \ |\ :4\]

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

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

\[x_{1} \cdot x_{2} = - 4;\]

\[x_{1} = 1 - не\ корень;\ \ \]

\[x_{2} = - 4\ (не\ подходит).\]

\[Ответ:нет\ корней.\]