Решебник по алгебре 9 класс Мерзляк Задание 750

 

\[\boxed{\mathbf{750\ (750).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[a_{1} = x^{2} - 4,\ \ a_{2} = 5x + 3,\ \ \]

\[a_{3} = 3x + 2,\ \ тогда\]

\[a_{2} = \frac{a_{1} + a_{3}}{2} \Longrightarrow \ \ 2a_{2} = a_{1} + a_{3}\]

\[2 \cdot (5x + 3) = x^{2} - 4 + 3x + 2\]

\[10x + 6 = x^{2} - 4 + 3x + 2\]

\[x^{2} - 7x - 8 = 0\]

\[x_{1} + x_{2} = 7,\ \ x_{1} = 8\]

\[x_{1}x_{2} = - 8,\ \ x_{2} = - 1\]

\[При\ x_{1} = 8 \Longrightarrow \ a_{1} =\]

\[= 64 - 4 = 60,\ \ \]

\[a_{2} = 40 + 3 = 43,\ \ \]

\[a_{3} = 24 + 2 = 26.\]

\[При\ x_{2} = - 1 \Longrightarrow a_{1} = 1 - 4 =\]

\[= - 3,\ \ a_{2} = - 5 + 3 = - 2,\]

\[\text{\ \ }a_{3} = - 3 + 2 = - 1.\]

\[Ответ:при\ x_{1} = - 1:\ \ \ \]

\[a_{1} = - 3,\ a_{2} = - 2,\ a_{3} = - 1;\ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ при\ x_{2} = 8:\ \ \]

\[\ a_{1} = 60,\ a_{2} = 43,\ a_{3} = 26.\]

\[\boxed{\mathbf{750.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[S = a^{2};\ \ где\ a - сторона\]

\[\ первого\ квадрата;\]

\[a_{1}^{2} = \left( \frac{a}{2} \right)^{2} + \left( \frac{a}{2} \right)^{2} = 2\frac{a^{2}}{4}\]

\[a_{2}^{2} = \frac{a^{2}}{2} \Longrightarrow \text{\ \ }a_{2} = \sqrt{\frac{a^{2}}{2}} =\]

\[= \frac{a}{\sqrt{2}} - второй\ квадрат;\]

\[S_{2} = a_{2}^{2} = \frac{a^{2}}{2};\ \]

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

\[= 2 \cdot \left( \frac{a}{2\sqrt{2}} \right)^{2}\text{\ \ }\]

\[a_{3}^{2} = 2\frac{a^{2}}{8}\text{\ \ }\]

\[a_{3}^{2} = \frac{a^{2}}{4} \Longrightarrow a_{3} = \frac{a}{2}\]

\[S_{3} = \frac{a^{2}}{4}\text{\ \ \ \ \ }и\ так\ далее.\]

\[Тогда\ \ a^{2},\frac{a^{2}}{2},\frac{a^{2}}{4},\ldots -\]

\[бесконечная\]

\[\ геометрическая\ прогрессия,\]

\[q = \frac{1}{2} \Longrightarrow \ S = \frac{b_{1}}{1 - q} =\]

\[= \frac{a^{2}}{1 - \frac{1}{2}} = \frac{a^{2}}{\frac{1}{2}} = 2a^{2}.\]

\[Ответ:S = 2a².\]