Решебник по алгебре 9 класс Макарычев Задание 599

 

\[\boxed{\text{599\ (599).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\left\{ \begin{matrix} 3x + y = 2\ \ \ \ \ \ \ \ \\ x^{2} - y^{2} = - 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 2 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - (2 - 3x)^{2} = - 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 2 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4 + 12x - 9x^{2} = - 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 2 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 8x^{2} - 12x - 8 = 0 \\ \end{matrix} \right.\ \]

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

\[D = 9 + 4 \cdot 2 \cdot 2 = 25\]

\[x_{1,2} = \frac{3 \pm 5}{4};\]

\[\left\{ \begin{matrix} x_{1} = 2\ \ \ \\ y_{1} = - 4 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x_{2} = - \frac{1}{2} \\ y_{2} = 3,5. \\ \end{matrix} \right.\ \]

\[Ответ:(2;\ - 4);\ \ ( - 0,5;3,5).\]

\[\boxed{\text{599.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ b_{1} = 125,\ \ \]

\[b_{3} = b_{1} \cdot q^{2} = 5,\ \ 125q² = 5,\]

\[q^{2} = \frac{1}{25},\ \ q = \pm \frac{1}{5},\ \ \]

\[b_{6} = 125 \cdot \left( \pm \frac{1}{5} \right)^{5} = \pm \frac{5^{3}}{5^{5}} =\]

\[= \pm \frac{1}{25};\ \]

\[\textbf{б)}\ b_{1} = - \frac{2}{9},\]

\[\text{\ \ }b_{3} = b_{1} \cdot q^{2} = - 2,\ \ \]

\[- \frac{2}{9q^{2}} = - 2,\]

\[q² = 9,\ \ q = \pm 3;\]

\[b_{7} = b_{1} \cdot q^{6} = - \frac{2}{9} \cdot ( \pm 3)^{6} =\]

\[= - \frac{2}{3^{2}} \cdot 3^{6} = - 2 \cdot 3^{4} = - 162;\]

\[\textbf{в)}\ b_{4} = b_{1} \cdot q³ = - 1,\]

\[\text{\ \ }b_{6} = b_{1} \cdot q^{5} = - 100,\ \ \]

\[\frac{b_{1}q^{5}}{b_{1}q^{3}} = \frac{- 100}{- 1} = 100;\]

\[q^{2} = 100,\ \ q = \pm 10;\]

\[b_{1} = \frac{b_{4}}{q^{3}} = \frac{- 1}{\pm 10³} = \pm 0,001.\]

\[Ответ:\ а)\ \frac{1}{25}\ \ или\ \left( - \frac{1}{25} \right);\ \]

\[\ б) - 162;\ \ в) - 0,001\ или\ 0,001.\]