Решебник по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 9

\[\boxed{\mathbf{9.}}\]

\[\textbf{а)}\ \sqrt{5x + 2} = x\sqrt{3}\]

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

\[3x^{2} - 5x - 2 = 0\]

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{5 + 7}{6} = 2;\]

\[x_{2} = \frac{5 - 7}{6} = - \frac{2}{6} = - \frac{1}{3}.\]

\[Проверка:\]

\[\sqrt{5 \cdot 2 + 2} = 2\sqrt{3}\]

\[\sqrt{12} = 2\sqrt{3}\]

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

\[x = 2 - корень.\]

\[\sqrt{- \frac{5}{3} + 2} = - \frac{1}{3}\sqrt{3}\]

\[\sqrt{\frac{1}{3}} \neq - \frac{1}{3}\sqrt{3}\]

\[x = - \frac{1}{3} - не\ является\ корнем.\]

\[Ответ:x = 2.\]

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

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

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

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{3 + 5}{4} = 2;\]

\[x_{2} = \frac{3 - 5}{4} = - \frac{1}{2}.\]

\[Проверка:\]

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

\[\sqrt{8} = \sqrt{8}\]

\[x = 2 - корень.\]

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

\[\sqrt{\frac{1}{2}} \neq - \frac{1}{2}\sqrt{2}\]

\[x = - \frac{1}{2} - не\ является\ корнем.\]

\[Ответ:x = 2.\]

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

\[2x + 5 = (x + 1)^{2}\]

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

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

\[x^{2} = 4\]

\[x = \pm 2.\]

\[Проверка:\]

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

\[\sqrt{9} = 3\]

\[3 = 3\]

\[x = 2 - корень.\]

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

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

\[1 \neq - 1\]

\[x = - 2 - не\ является\ корнем.\]

\[Ответ:x = 2.\]

\[\textbf{г)}\ \sqrt{3x + 7} = 2x + 3\]

\[3x + 7 = (2x + 3)^{2}\]

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

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

\[D = 81 - 32 = 49\]

\[x_{1} = \frac{- 9 + 7}{8} = - \frac{2}{8} = - \frac{1}{4};\]

\[x_{2} = \frac{- 9 - 7}{8} = - 2.\]

\[Проверка:\]

\[\sqrt{3 \cdot \left( - \frac{1}{4} \right) + 7} = 2 \cdot \left( - \frac{1}{4} \right) + 3\]

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

\[\sqrt{\frac{25}{4}} = \frac{5}{2}\]

\[\frac{5}{2} = \frac{5}{2}\]

\[x = - \frac{1}{4} - корень.\]

\[\sqrt{3 \cdot ( - 2) + 7} = 2 \cdot ( - 2) + 3\]

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

\[1 \neq - 1\]

\[x = - 2 - не\ является\ корнем.\]

\[Ответ:x = - \frac{1}{4}.\]

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

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

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

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

\[x(x - 6) = 0\]

\[x = 0;\ \ x = 6.\]

\[Проверим:\]

\[\sqrt{2 \cdot 0 - 4 \cdot 0 + 1} = 0 + 1\]

\[\sqrt{1} = 1\]

\[1 = 1\]

\[x = 0 - корень.\]

\[\sqrt{2 \cdot 36 - 4 \cdot 6 + 1} = 6 + 1\]

\[\sqrt{49} = 7\]

\[7 = 7\]

\[x = 6 - корень.\]

\[Ответ:x = 0;x = 6.\]

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

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

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

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

\[2x(x - 1) = 0\]

\[x = 0;\ \ x = 1.\]

\[Проверка:\]

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

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

\[1 \neq - 1\]

\[x = 0 - не\ является\ корнем.\]

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

\[\sqrt{0} = 0\]

\[0 = 0\]

\[x = 1 - корень.\]

\[Ответ:x = 1.\]