ГДЗ по алгебре 8 класс Макарычев Задание 639

\[\boxed{\text{639\ (639).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

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

\[уравнение.\ \]

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

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

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

=\(0\)

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

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

\[x^{2} - 4x + 1 = 0 - искомое\ \]

\[уравнение.\ \]