ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 670

\[\boxed{\text{670.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ x_{1} = \frac{\sqrt{3} - 1}{2};\ \ \ \ x_{2} = \frac{\sqrt{3} + 1}{2}\]

\[\left( x - x_{1} \right)\left( x - x_{2} \right) = 0\]

\[\left( x - \frac{\sqrt{3} - 1}{2} \right)\left( x - \frac{\sqrt{3} + 1}{2} \right) = 0\]

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

\[+ \frac{\sqrt{3} - 1}{2} \cdot \frac{\sqrt{3} + 1}{2} = 0\]

\[x^{2} - x \cdot \left( \frac{\sqrt{3} + 1}{2} + \frac{\sqrt{3} - 1}{2} \right) +\]

\[+ \frac{\left( \sqrt{3} - 1 \right)\left( \sqrt{3} + 1 \right)}{4} = 0\]

\[x^{2} - x \cdot \left( \frac{\sqrt{3} + 1 + \sqrt{3} - 1}{2} \right) +\]

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

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

\[x^{2} - \sqrt{3}x + \frac{1}{2} = 0;\ \ \ 2x^{2} -\]

\[- \sqrt{3x} + 1 = 0 - искомое\ \]

\[уравнение.\ \]

\[\textbf{б)}\ x_{1} = 2 - \sqrt{3};\ \ x_{2} = \frac{1}{2 - \sqrt{3}}\]

\[\left( x - x_{1} \right)\left( x - x_{2} \right) = 0\]

\[\left( x - \left( 2 - \sqrt{3} \right) \right)\left( x - \frac{1}{2 - \sqrt{3}} \right) = 0\]

\[x^{2} - \frac{1}{2 - \sqrt{3}}x - \left( 2 - \sqrt{3} \right)x +\]

\[+ \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = 0\]

\[x^{2} - x \cdot \left( \frac{1}{2 - \sqrt{3}} + \left( 2 - \sqrt{3} \right) \right) +\]

\[+ 1 = 0\]

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

\[+ 1 = 0\]

\[x^{2} - \frac{8 - 4\sqrt{3}}{2 - \sqrt{3}}x + 1 = 0\]

\[x^{2} - \frac{4 \cdot \left( 2 - \sqrt{3} \right)}{2 - \sqrt{3}}x + 1 = 0\]

\[x^{2} - 4x + 1 = 0 - искомое\]

\[\ уравнение.\ \]