Решебник по алгебре 8 класс Мерзляк ФГОС Задание 774

 

\[\boxed{\mathbf{774\ (774).\ }Еуроки - \ ДЗ\ без\ мороки}\]

\[P_{1} = \left( a_{1} + b_{1} \right) \cdot 2,\ \ \]

\[P_{1} = 1000\]

\[\left( a_{1} + b_{1} \right) \cdot 2 = 1000\]

\[a_{1} + b_{1} = 500\]

\[a_{1}\ может\ быть\ от\ 1\ до\ 499\ \ \ и\ \ \ \ \ \]

\[b_{1}\ от\ 1\ до\ 499.\]

\[P_{2} = \left( a_{2} + b_{2} \right) \cdot 2,\ \ P_{2} = 1002\]

\[\left( a_{2} + b_{2} \right) \cdot 2 = 1002\]

\[a_{2} + b_{2} = 501\]

\[a_{2}\ \ может\ быть\ от\ 1\ до\ 500\ \ \ и\ \ \]

\[b_{2}\ \ от\ 1\ до\ 500.\]

\[Следовательно,\ больше\ \]

\[прямоугольников\ \]

\[с\ периметром\ 1002.\]

\[\boxed{\mathbf{7}\mathbf{7}\mathbf{4}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[x_{1}² + x_{2}² = \frac{46}{9}\]

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

\[= \left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2} = \frac{46}{9}\]

\[3x^{2} + ax - 7 = 0\]

\[x_{1}x_{2} = - \frac{7}{3}\]

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

\[\left( - \frac{a}{3} \right)^{2} + \frac{2 \cdot 7}{3} = \frac{46}{9}\]

\[\frac{a^{2}}{9} + \frac{14}{3} = \frac{46}{9}\]

\[\frac{a^{2}}{9} = \frac{46}{9} - \frac{14}{3}\]

\[\frac{a^{2}}{9} = \frac{4}{9}\]

\[a^{2} = 4\]

\[a = 2;\ \ \ \ a = - 2\]

\[Ответ:\ a = - 2\ \ или\ a = \ 2.\]