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

\[\boxed{\mathbf{8.}}\]

\[\textbf{а)}\ \sqrt{x^{2} - 4x + 1} = \sqrt{3x + 1}\]

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

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

\[x(x - 7) = 0\]

\[x = 0;\ \ x = 7.\]

\[Проверка:\]

\[\sqrt{0 - 4 \cdot 0 + 1} = \sqrt{3 \cdot 0 + 1}\]

\[\sqrt{1} = \sqrt{1}\]

\[x = 0 - корень.\]

\[\sqrt{7^{2} - 4 \cdot 7 + 1} = \sqrt{3 \cdot 7 + 1}\]

\[\sqrt{22} = \sqrt{22}\]

\[x = 7 - корень.\]

\[Ответ:x = 1;\ \ x = 7.\]

\[\textbf{б)}\ \sqrt{2x^{2} - 4x + 5} =\]

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

\[2x^{2} - 4x + 5 = 3x^{2} - x + 1\]

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

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 4\]

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

\[Проверка:\]

\[\sqrt{2 \cdot 16 + 16 + 5} =\]

\[= \sqrt{3 \cdot 16 + 4 + 1}\]

\[\sqrt{53} = \sqrt{53}\]

\[x = - 4 - корень.\]

\[\sqrt{2 \cdot 1 - 4 \cdot 1 + 5} =\]

\[= \sqrt{3 \cdot 1 - 1 + 1}\]

\[\sqrt{3} = \sqrt{3}\]

\[x = 1 - корень.\]

\[Ответ:x = - 4;\ \ x = 1.\]

\[\textbf{в)}\ \sqrt{x^{2} - 3x} = \sqrt{4x - 10}\]

\[x^{2} - 3x = 4x - 10\]

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

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

\[x_{1} = 5;\ \ x_{2} = 2.\]

\[Проверка:\]

\[\sqrt{25 - 15} = \sqrt{20 - 10}\]

\[\sqrt{10} = \sqrt{10}\]

\[x = 5 - \ является\ корнем.\]

\[\sqrt{4 - 6} = \sqrt{8 - 10}\]

\[x = 2 - не\ корень.\]

\[Ответ:x = 5.\]

\[\textbf{г)}\ \sqrt{x^{2} - 3x - 3} =\]

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

\[x^{2} - 3x - 3 = 2x^{2} - 2x - 9\]

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

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

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

\[Проверка:\]

\[\sqrt{9 + 9 - 3} = \sqrt{18 + 6 - 9}\]

\[\sqrt{15} = \sqrt{15}\]

\[x = - 3 - корень.\]

\[\sqrt{4 - 6 - 3} = \sqrt{8 - 4 - 9}\]

\[x = 2 - не\ корень.\]

\[Ответ:x = - 3.\]