Решебник по алгебре 9 класс Мерзляк Задание 752

 

\[\boxed{\mathbf{752\ (752).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

\[\text{\ \ }a_{3} = 4y + 13,\ \ \]

\[a_{4} = 2y^{2} - y + 25\]

\[a_{1} + a_{4} = a_{3} + a_{2}\]

\[y^{2} - 2y + 2y^{2} - y + 25 =\]

\[= 3y + 5 + 4y + 13\]

\[3y^{2} - 10y + 7 = 0\]

\[D = 100 - 84 = 16\]

\[y_{1} = \frac{10 + 4}{6} = \frac{7}{3},\]

\[\text{\ \ }y_{2} = \frac{10 - 4}{6} = 1\]

\[При\ y = \frac{7}{3},\ \ \]

\[a_{1} = \frac{49}{9} - \frac{14}{3} = \frac{7}{9},\]

\[\text{\ \ }a_{2} = 7 + 5 = 12,\ \]

\[a_{3} = \frac{28}{3} + 13 = \frac{67}{3},\]

\[\text{\ \ }a_{4} = \frac{2 \cdot 49}{9} - \frac{7}{3} + 25 = \frac{302}{9}\]

\[Проверим:a_{2} - a_{1} \neq a_{3} -\]

\[- a_{2} \neq a_{4} - a_{3},\ \ \]

\[следовательно,\ y = \frac{7}{3}\ \Longrightarrow не\ \]

\[удовлетворяет.\]

\[При\ y = 1:\ \ \]

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

\[a_{2} = 3 + 5 = 8,\ \ \]

\[a_{3} = 4 + 13 = 17,\]

\[a_{4} = 2 - 1 + 25 = 26.\]

\[Ответ:при\ y = 1 \Longrightarrow \ a_{1} =\]

\[= - 1,a_{2} = 8,\ a_{3} = 17,\ a_{4} = 26.\ \]

\[\boxed{\mathbf{752.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[В\ квадрат\ со\ стороной\ a\ \]

\[вписана\ окружность\ R = \frac{a}{2};\ \]

\[длина\ окружности\ \]

\[c_{1} = 2\pi\frac{a}{2} = \pi a;\ \ S_{1}` = \pi\left( \frac{a}{2} \right)^{2} =\]

\[= \frac{\pi a^{2}}{4} - площадь\ круга;\]

\[\text{\ \ }P_{1} = 4a -\]

\[периметр\ квадрата;\ \ S_{1} =\]

\[= a^{2} - площадь\ квадрата.\]

\[В\ окружность\ вписан\ квадрат\ \]

\[со\ стороной\ a = \frac{a}{2}\ :\frac{\sqrt{2}}{2} =\]

\[= \frac{2a}{2\sqrt{2}} = \frac{a\sqrt{2}}{2};\]

\[P_{2} = \frac{4a\sqrt{2}}{2} = 2a\sqrt{2};\ \]

\[\ S_{2} = \left( \frac{a\sqrt{2}}{2} \right)^{2} = \frac{a^{2} \cdot 2}{4} = \frac{a^{2}}{2}.\]

\[В\ квадрат\ вписана\ окружность\]

\[\text{\ \ }R = \frac{a_{2}}{2} = \frac{a\sqrt{2}}{2 \cdot 2} = \frac{a\sqrt{2}}{4};\ \]

\[c_{2} = 2\pi \cdot \frac{a\sqrt{2}}{4} = \frac{\text{aπ}\sqrt{2}}{2};\]

\[S_{2} = \pi \cdot \left( \frac{a\sqrt{2}}{4} \right)^{2} = \frac{2\pi a^{2}}{16} = \frac{\pi a^{2}}{8}.\]

\[1)\ 4a,\ 2a\sqrt{2},2a,\ldots -\]

\[геометрическая\ прогрессия;\ \]

\[\ q = \frac{\sqrt{2}}{2},\ \]

\[S = \frac{b_{1}}{1 - q} = \frac{4a}{1 - \frac{\sqrt{2}}{2}} = \frac{4a}{\frac{2 - \sqrt{2}}{2}} =\]

\[= \frac{8a}{2 - \sqrt{2}} = \frac{8a\left( 2 + \sqrt{2} \right)}{\left( 2 - \sqrt{2} \right)\left( 2 + \sqrt{2} \right)} =\]

\[= \frac{8a\left( 2 + \sqrt{2} \right)}{4 - 2} =\]

\[= 4a\left( 2 + \sqrt{2} \right);\]

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

\[геометрическая\ прогрессия;\]

\[\text{\ \ }q = \frac{1}{2},\]

\[S = \frac{b_{1}}{1 - q} = \frac{a^{2}}{1 - \frac{1}{2}} = \frac{a^{2}}{\frac{1}{2}} = 2a^{2};\]

\[3)\ \pi a,\frac{\text{aπ}\sqrt{2}}{2},\ \ldots -\]

\[геометрическая\ прогрессия;\]

\[\text{\ \ }q = \frac{\sqrt{2}}{2},\]

\[S = \frac{\text{πa}}{1 - \frac{\sqrt{2}}{2}} = \frac{\text{πa}}{\frac{2 - \sqrt{2}}{2}} =\]

\[= \frac{2\pi a}{2 - \sqrt{2}} = \frac{2\pi a\left( 2 + \sqrt{2} \right)}{\left( 2 - \sqrt{2} \right)\left( 2 + \sqrt{2} \right)} =\]

\[= \frac{2\pi a\left( 2 + \sqrt{2} \right)}{4 - 2} =\]

\[= \pi a\left( 2 + \sqrt{2} \right);\]

\[4)\frac{\pi a^{2}}{4},\frac{\pi a^{2}}{8},\ \ldots -\]

\[геометрическая\ прогрессия;\ \ \]

\[q = \frac{1}{2},\]

\[S = \frac{\pi a^{2}}{4 \cdot \left( 1 - \frac{1}{2} \right)} = \frac{\pi a^{2}}{4 \cdot \frac{1}{2}} = \frac{\pi a^{2}}{2}.\]

\[Ответ:1)\ 4a\left( 2 + \sqrt{2} \right);\ 2)\ 2a²;\ \ \]

\[3)\ \text{πa}\left( 2 + \sqrt{2} \right);\ \ 4)\ \frac{\pi a^{2}}{2}.\]