ГДЗ по алгебре 9 класс Макарычев Задание 435

\[\boxed{\text{435.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} 2x + 4y = 5 \cdot (x - y) \\ x^{2} - y^{2} = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 2x + 4y = 5x - 5y \\ x^{2} - y^{2} = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 3x = 9y\ \ \ \ \ \ \\ x^{2} - y^{2} = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 3y\ \ \ \ \ \ \ \ \ \ \ \ \\ 9y^{2} - y^{2} = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 3y\ \ \\ 8y^{2} = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 3y \\ y^{2} = \frac{3}{4} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\left\{ \begin{matrix} y_{1} = \frac{\sqrt{3}}{2}\ \\ x_{1} = \frac{3\sqrt{3}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \]

\[\left\{ \begin{matrix} y_{2} = - \frac{\sqrt{3}}{2}\text{\ \ \ \ } \\ x_{2} = - \frac{3\sqrt{3}}{2}. \\ \end{matrix} \right.\ \]

\[\textbf{б)}\ \left\{ \begin{matrix} u - v = 6 \cdot (u + v) \\ u^{2} - v^{2} = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} - 5u = 7v\ \ \ \ \\ u^{2} - v^{2} = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} u = - \frac{7}{5}\text{v\ \ \ \ \ \ \ \ \ } \\ \frac{49}{25}v^{2} - v^{2} = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} u = - \frac{7}{5}v \\ v^{2} = \frac{25}{4}\text{\ \ \ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} v_{1} = 2,5\ \ \ \ \ \\ u_{1} = - 3,5 \\ \end{matrix} \right.\ \text{\ \ }или\ \ \ \]

\[\left\{ \begin{matrix} v_{2} = - 2,5 \\ u_{2} = 3,5\ \ \ \\ \end{matrix} \right.\ .\]

\[Ответ:а)\ \left( \frac{3\sqrt{3}}{2};\frac{\sqrt{3}}{2} \right);\ \ \]

\[\left( - \frac{3\sqrt{3}}{2};\ - \frac{\sqrt{3}}{2} \right);\ \ \ \]

\[\textbf{б)}\ ( - 3,5;2,5);\ \ (3,5;\ - 2,5).\]