Решебник по алгебре 8 класс Макарычев ФГОС Задание 646

\[\boxed{\text{646\ (646).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x^{2} - ax + a - 3 = 0\]

\[по\ теореме\ Виета:x_{1} + x_{2} = a;\]

\[\text{\ \ }x_{1}x_{2} = a - 3\]

\[\left( x_{1} + x_{2} \right)^{2} = (a)^{2}\]

\[x_{1}^{2} + 2x_{1}x_{2} + x_{2}^{2} = a^{2}\]

\[x_{1}^{2} + x_{2}^{2} + 2 \cdot (a - 3) = a^{2}\]

\[x_{1}^{2} + x_{2}^{2} = a^{2} - 2a + 6 =\]

\[= a^{2} - 2a + 1 + 5 =\]

\[= (a - 1)^{2} + 5,\]

\[то\ есть\ x_{1}^{2} + x_{2}^{2} - принимает\ \]

\[наименьшее\ значение\ при\ \]

\[a = 1\]

\[x_{1}^{2} + x_{2}^{2} = 1^{2} - 2 \cdot 1 + 6 =\]

\[= 1 - 2 + 6 = 5\]

\[Ответ:x_{1}^{2} + x_{2}^{2} = 5\ при\ a = 1\text{.\ \ }\]

\[\boxed{\text{646.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \frac{x - y}{\sqrt{x} - \sqrt{y}} - \sqrt{x} =\]

\[= \frac{\left( \sqrt{x} - \sqrt{y} \right)\left( \sqrt{x} + \sqrt{y} \right)}{\left( \sqrt{x} - \sqrt{y} \right)} -\]

\[- \sqrt{x} = \sqrt{x} + \sqrt{y} - \sqrt{x} = \sqrt{y}\]

\[\textbf{б)}\ \sqrt{x} - \frac{x - y}{\sqrt{x} + \sqrt{y}} = \sqrt{x} -\]

\[- \frac{\left( \sqrt{x} - \sqrt{y} \right)\left( \sqrt{x} + \sqrt{y} \right)}{\left( \sqrt{x} + \sqrt{y} \right)} =\]

\[= \sqrt{x} - \sqrt{x} + \sqrt{y} = \sqrt{y}\ \]