Решебник по алгебре 9 класс Макарычев Задание 712

 

\[\boxed{\text{712\ (712).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[S_{5} = \frac{11}{64},\ \ S_{10} - S_{5} = - 5\frac{1}{2},\]

\[S_{5} = x_{1} \cdot \frac{q^{5} - 1}{q - 1},\ \ \]

\[S_{10} = x_{1} \cdot \frac{q^{10} - 1}{q - 1},\]

\[S_{10} - S_{5} =\]

\[= \frac{x_{1}}{q - 1} \cdot \left( q^{10} - 1 - q^{5} + 1 \right) =\]

\[= \frac{x_{1} \cdot q^{5} \cdot \left( q^{5} - 1 \right)}{q - 1} = q^{6} \cdot S_{5},\]

\[q^{5} = \frac{S_{10} - S_{5}}{S_{5}} = - 5\frac{1}{2} \cdot \frac{64}{11} = - 32,\]

\[q = - 2,\]

\[S_{15} - S_{10} =\]

\[= x_{1} \cdot \frac{q^{15} - 1}{q - 1} - x_{1} \cdot \ \frac{q^{10} - 1}{q - 1} =\]

\[= \frac{x_{1}}{q - 1} \cdot \left( q^{15} - 1 - q^{10} + 1 \right) =\]

\[= \frac{x_{1} \cdot q^{10} \cdot \left( q^{5} - 1 \right)}{q - 1} = q^{10} \cdot S_{5} =\]

\[= ( - 2)^{10} \cdot \frac{11}{64} = 176.\]

\[\boxed{\text{712.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ a + b = 6,\ \ ab = 3:\]

\[a^{2} + b^{2} = a^{2} + 2ab + b^{2} - 2ab =\]

\[= (a + b)^{2} - 2ab = 6^{2} - 2 \cdot 3 =\]

\[= 30.\]

\[\textbf{б)}\ c + \frac{1}{c} = 2,5:\]

\[c^{2} + \frac{1}{c^{2}} = c^{2} + 2 + \frac{1}{c^{2}} - 2 =\]

\[= \left( c + \frac{1}{c} \right)^{2} - 2 = {2,5}^{2} - 2 =\]

\[= 4,25.\ \]