ГДЗ по алгебре 8 класс Макарычев Задание 637

\[\boxed{\text{637\ (637).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ при\ x = 5 + 2\sqrt{6},\]

\[\ \ y = 5 - 2\sqrt{6}:\]

\[\frac{\text{xy}}{x + y} = \frac{\left( 5 + 2\sqrt{6} \right)\left( 5 - 2\sqrt{6} \right)}{5 + 2\sqrt{6} + 5 - 2\sqrt{6}} =\]

\[= \frac{25 - 4 \cdot 6}{10} = \frac{25 - 24}{10} =\]

\[= \frac{1}{10} = 0,1.\]

\[\textbf{б)}\ при\ x = \sqrt{11} + \sqrt{3},\]

\[\ \ y = \sqrt{11} - \sqrt{3}\]

\[\frac{x^{2} + y^{2}}{\text{xy}} =\]

\[= \frac{\left( \sqrt{11} + \sqrt{3} \right)^{2} + \left( \sqrt{11} - \sqrt{3} \right)^{2}}{\left( \sqrt{11} + \sqrt{3} \right)\left( \sqrt{11} - \sqrt{3} \right)} =\]

\[= \frac{28}{8} = \frac{14}{4} = \frac{7}{2} = 3,5.\ \]