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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 9
Задание 9
\[\boxed{\mathbf{9.}}\]
\[\textbf{а)}\ \sqrt{5x + 2} = x\sqrt{3}\]
\[5x + 2 = 3x^{2}\]
\[3x^{2} - 5x - 2 = 0\]
\[D = 25 + 24 = 49\]
\[x_{1} = \frac{5 + 7}{6} = 2;\]
\[x_{2} = \frac{5 - 7}{6} = - \frac{2}{6} = - \frac{1}{3}.\]
\[Проверка:\]
\[\sqrt{5 \cdot 2 + 2} = 2\sqrt{3}\]
\[\sqrt{12} = 2\sqrt{3}\]
\[2\sqrt{3} = 2\sqrt{3}\]
\[x = 2 - корень.\]
\[\sqrt{- \frac{5}{3} + 2} = - \frac{1}{3}\sqrt{3}\]
\[\sqrt{\frac{1}{3}} \neq - \frac{1}{3}\sqrt{3}\]
\[x = - \frac{1}{3} - не\ является\ корнем.\]
\[Ответ:x = 2.\]
\[\textbf{б)}\ \sqrt{3x + 2} = x\sqrt{2}\]
\[3x + 2 = 2x^{2}\]
\[2x^{2} - 3x - 2 = 0\]
\[D = 9 + 16 = 25\]
\[x_{1} = \frac{3 + 5}{4} = 2;\]
\[x_{2} = \frac{3 - 5}{4} = - \frac{1}{2}.\]
\[Проверка:\]
\[\sqrt{3 \cdot 2 + 2} = 2\sqrt{2}\]
\[\sqrt{8} = \sqrt{8}\]
\[x = 2 - корень.\]
\[\sqrt{- \frac{3}{2} + 2} = - \frac{1}{2}\sqrt{2}\]
\[\sqrt{\frac{1}{2}} \neq - \frac{1}{2}\sqrt{2}\]
\[x = - \frac{1}{2} - не\ является\ корнем.\]
\[Ответ:x = 2.\]
\[\textbf{в)}\ \sqrt{2x + 5} = x + 1\]
\[2x + 5 = (x + 1)^{2}\]
\[2x + 5 = x^{2} + 2x + 1\]
\[x^{2} - 4 = 0\]
\[x^{2} = 4\]
\[x = \pm 2.\]
\[Проверка:\]
\[\sqrt{2 \cdot 2 + 5} = 2 + 1\]
\[\sqrt{9} = 3\]
\[3 = 3\]
\[x = 2 - корень.\]
\[\sqrt{2 \cdot ( - 2) + 5} = - 2 + 1\]
\[\sqrt{1} = - 1\]
\[1 \neq - 1\]
\[x = - 2 - не\ является\ корнем.\]
\[Ответ:x = 2.\]
\[\textbf{г)}\ \sqrt{3x + 7} = 2x + 3\]
\[3x + 7 = (2x + 3)^{2}\]
\[3x + 7 = 4x^{2} + 12x + 9\]
\[4x^{2} + 9x + 2 = 0\]
\[D = 81 - 32 = 49\]
\[x_{1} = \frac{- 9 + 7}{8} = - \frac{2}{8} = - \frac{1}{4};\]
\[x_{2} = \frac{- 9 - 7}{8} = - 2.\]
\[Проверка:\]
\[\sqrt{3 \cdot \left( - \frac{1}{4} \right) + 7} = 2 \cdot \left( - \frac{1}{4} \right) + 3\]
\[\sqrt{- \frac{3}{4} + 7} = - \frac{1}{2} + 3\]
\[\sqrt{\frac{25}{4}} = \frac{5}{2}\]
\[\frac{5}{2} = \frac{5}{2}\]
\[x = - \frac{1}{4} - корень.\]
\[\sqrt{3 \cdot ( - 2) + 7} = 2 \cdot ( - 2) + 3\]
\[\sqrt{1} = - 1\]
\[1 \neq - 1\]
\[x = - 2 - не\ является\ корнем.\]
\[Ответ:x = - \frac{1}{4}.\]
\[\textbf{д)}\ \sqrt{2x^{2} - 4x + 1} = x + 1\]
\[2x^{2} - 4x + 1 = (x + 1)^{2}\]
\[2x^{2} - 4x + 1 = x^{2} + 2x + 1\]
\[x^{2} - 6x = 0\]
\[x(x - 6) = 0\]
\[x = 0;\ \ x = 6.\]
\[Проверим:\]
\[\sqrt{2 \cdot 0 - 4 \cdot 0 + 1} = 0 + 1\]
\[\sqrt{1} = 1\]
\[1 = 1\]
\[x = 0 - корень.\]
\[\sqrt{2 \cdot 36 - 4 \cdot 6 + 1} = 6 + 1\]
\[\sqrt{49} = 7\]
\[7 = 7\]
\[x = 6 - корень.\]
\[Ответ:x = 0;x = 6.\]
\[\textbf{е)}\ \sqrt{3x^{2} - 4x + 1} = x - 1\]
\[3x^{2} - 4x + 1 = (x - 1)^{2}\]
\[3x^{2} - 4x + 1 = x^{2} - 2x + 1\]
\[2x^{2} - 2x = 0\]
\[2x(x - 1) = 0\]
\[x = 0;\ \ x = 1.\]
\[Проверка:\]
\[\sqrt{3 \cdot 0 - 4 \cdot 0 + 1} = 0 - 1\]
\[\sqrt{1} = - 1\]
\[1 \neq - 1\]
\[x = 0 - не\ является\ корнем.\]
\[\sqrt{3 \cdot 1 - 4 \cdot 1 + 1} = 1 - 1\]
\[\sqrt{0} = 0\]
\[0 = 0\]
\[x = 1 - корень.\]
\[Ответ:x = 1.\]