Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 7

\[\boxed{\mathbf{7.}}\]

\[\textbf{а)}\ \sqrt{x} = x - 2\]

\[x = (x - 2)^{2}\]

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

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

\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = 4\]

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

\[Проверка:\]

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

\[1 = - 1\]

\[x = 1 - не\ является\ корнем.\]

\[\sqrt{4} = 4 - 2\]

\[2 = 2\]

\[x = 4 - корень\ уравнения.\]

\[Ответ:x = 4.\]

\[\textbf{б)}\ \sqrt{3x} = 2x - 3\]

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

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

\[4x^{2} - 15x + 9 = 0\]

\[D = 225 - 144 = 81\]

\[x_{1} = \frac{15 + 9}{8} = 3;\]

\[x_{2} = \frac{15 - 9}{8} = \frac{6}{8} = \frac{3}{4}.\]

\[Проверка:\]

\[\sqrt{3 \cdot 3} = 2 \cdot 3 - 3\]

\[3 = 3\]

\[x = 3 - является\ корнем.\]

\[\sqrt{3 \cdot \frac{3}{4}} = 2 \cdot \frac{3}{4} - 3\]

\[\frac{3}{2} = \frac{1}{2} - 3\]

\[\frac{3}{2} = - \frac{3}{2}\]

\[x = \frac{3}{4} - не\ является\ корнем.\]

\[Ответ:x = 3.\]

\[\textbf{в)}\ \sqrt{2x - 1} = x\]

\[2x - 1 = x^{2}\]

\[x^{2} - 2x + 1 = 0\]

\[(x - 1)^{2} = 0\]

\[x - 1 = 0\]

\[x = 1.\]

\[Проверка:\]

\[\sqrt{2 \cdot 1 - 1} = 1\]

\[1 = 1\]

\[x = 1 - корень.\]

\[Ответ:x = 1.\]

\[\textbf{г)}\ \sqrt{3x - 2} = x\]

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

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

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

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

\[Проверка:\]

\[\sqrt{3 \cdot 1 - 2} = 1\]

\[1 = 1\]

\[x = 1 - корень.\]

\[\sqrt{3 \cdot 2 - 2} = 2\]

\[\sqrt{4} = 2\]

\[x = 2 - корень.\]

\[Ответ:x = 1;x = 2.\]