ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 668

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

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

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

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

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

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

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

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

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