ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 711

\[\boxed{\text{711.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x - y = 5 \\ \frac{1}{x} + \frac{1}{y} = \frac{1}{6} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 5\ \ \ \ \ \ \ \ \\ \frac{1}{y + 5} + \frac{1}{y} = \frac{1}{6} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{6y + 6 \cdot (y + 5) - y(y + 5)}{6y(y + 5)} = 0 \\ \end{matrix} \right.\ \]

\[6y + 6y + 30 - y^{2} - 5y = 0\]

\[y^{2} - 7y - 30 = 0\]

\[D = 7^{2} + 4 \cdot 30 = 169\]

\[y_{1} = \frac{7 + 13}{2} = 10;\]

\[y_{2} = \frac{7 - 13}{2} = - 3.\]

\[1)\ y_{1} = 10;\ \ \ \ \ x_{1} = 15;\]

\[2)\ y_{2} = - 3;\ \ \ \ x_{2} = 2.\ \ \]

\[Ответ:(2; - 3);(15;10).\]

\[\textbf{б)}\ \left\{ \begin{matrix} x + y = 6 \\ \frac{1}{x} - \frac{1}{y} = \frac{1}{4} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 6 - x\ \ \ \ \ \\ \frac{1}{x} - \frac{1}{6 - x} = \frac{1}{4} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 6 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{4 \cdot (6 - x) - 4x - x(6 - x)}{4x(6 - x)} = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 6 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{24 - 4x - 4x - 6x + x²}{4x(6 - x)} = 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 14x + 24 = 0\]

\[D_{1} = 49 - 24 = 25\]

\[x_{1} = 7 - 5 = 2;\ \ \ \]

\[x_{2} = 7 + 5 = 12.\]

\[1)\ x_{1} = 2;\ \ \ \ \ \ \ y_{1} = 4;\]

\[2)\ x_{2} = 12;\ \ \ \ y_{2} = - 6.\]

\[Ответ:(2;4);(12; - 6).\]

\[\textbf{в)}\ \left\{ \begin{matrix} 3x + y = 1\ \ \ \\ \frac{1}{x} + \frac{1}{y} = - 2,5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 1 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{1}{x} + \frac{1}{1 - 3x} = - 2,5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 1 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{2 - 6x + 2x + 5x - 15x²}{x(1 - 3x)} = 0 \\ \end{matrix} \right.\ \]

\[- 15x^{2} + x + 2 = 0\]

\[15x^{2} - x - 2 = 0\]

\[D = 1 + 120 = 121\]

\[x_{1} = \frac{1 - 11}{30} = - \frac{1}{3};\ \ \]

\[x_{2} = \frac{1 + 11}{30} = \frac{12}{30} = \frac{2}{5}.\]

\[1)\ x_{1} = - \frac{1}{3};\ \ \ \ y_{1} = 2;\]

\[2)\ x_{2} = \frac{2}{5};\ \ \ \ \ \ \ y_{2} = - \frac{1}{5}.\]

\[Ответ:\left( - \frac{1}{3};2 \right);\ \ \left( \frac{2}{5}; - \frac{1}{5} \right).\]

\[\textbf{г)}\ \left\{ \begin{matrix} \frac{1}{y} - \frac{1}{x} = \frac{1}{3}\text{\ \ } \\ x - 2y = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 2y + 3\ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{3x - 3y - xy}{3xy} = 0 \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} x = 2y + 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3 \cdot (2y + 3) - 3y - y(2y + 3) = 0 \\ \end{matrix} \right.\ \]

\[2y^{2} - y - 6 = 0\]

\[D = 1 + 4 \cdot 2 \cdot 6 = 49\]

\[y_{1} = \frac{1 + 7}{4} = 2;\ \ \ \]

\[y_{2} = \frac{1 - 7}{4} = - 1,5.\]

\[1)\ y_{1} = 2;\ \ x_{1} = 6;\]

\[2)\ y_{2} = - 1,5;\ \ x_{2} = - 1.\]

\[Ответ:( - 1; - 1,5);\ \ (6;2).\]