Решебник по алгебре 9 класс Макарычев Задание 501

 

\[\boxed{\text{501\ (}\text{c}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x - y \leq 0 \\ x + y \leq 5 \\ x \geq 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Прямоугольный\ треугольник.\]

\[S = \frac{1}{2} \cdot 5 \cdot 2,5 = 6,25\ ед².\]

\[\textbf{б)}\ \left\{ \begin{matrix} - 2 \leq x \leq 4 \\ - 1 \leq y \leq 1 \\ \end{matrix} \right.\ \]

\[Прямоугольник.\]

\[S = 6 \cdot 2 = 12\ ед².\]

\[\boxed{\text{501.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + y^{2} + x + y = 18 \\ x^{2} - y^{2} + x - y = 6\ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 2x^{2} + 2x = 24 \\ 2y^{2} + 2y = 12 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} + x - 12 = 0 \\ y^{2} + y - 6 = 0\ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - 3)(x + 4) = 0 \\ (y - 2)(y + 3) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

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

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

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2}y^{2} + xy = 72 \\ x + y = 6\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Пусть\ xy = t,\ тогда\ \]

\[\ t^{2} + t - 72 = 0 \Longrightarrow t_{1,2} = - 9;8;\]

\[1)\ \left\{ \begin{matrix} xy = - 9\ \ \\ x = 6 - y \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y(6 - y) = - 9 \\ x = 6 - y\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y^{2} - 6y - 9 = 0 \\ x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[D = 9 + 9 = 18\]

\[y_{1,2} = 3 \pm 3\sqrt{2}\]

\[2)\ \left\{ \begin{matrix} xy = 8\ \ \ \\ x = 6 - y \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y^{2} - 6y + 8 = 0 \\ x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 4 \\ x_{1} = 2 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \left\{ \begin{matrix} y_{2} = 2\ \\ x_{2} = 4. \\ \end{matrix} \right.\ \]

\[\textbf{в)}\ \left\{ \begin{matrix} (x + y)^{2} - 2 \cdot (x + y) = 15 \\ x + xy + y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Пусть\ t = x + y,\ \ тогда\ \ \]

\[t^{2} - 2t - 15 = 0,\]

\[t_{1,2} = - 3;5;\ \]

\[1)\ \left\{ \begin{matrix} x + y = - 3 \\ xy = 11 + 3 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = - y - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - y^{2} - 3y - 14 = 0 \\ \end{matrix} \right.\ \]

\[y^{2} + 3y + 14 = 0\]

\[D = 9 - 4 \cdot 14 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет;\]

\[2)\ \left\{ \begin{matrix} x + y = 5\ \ \ \ \\ xy + 5 = 11 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - y^{2} + 5y - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[\textbf{г)}\ \left\{ \begin{matrix} (x + y)^{2} - 4 \cdot (x + y) = 45 \\ (x - y)^{2} - 2 \cdot (x - y) = 3\ \ \\ \end{matrix} \right.\ \]

\[Пусть\ a = x + y,\ \ b = x - y,\]

\[\left\{ \begin{matrix} a^{2} - 4a - 45 = 0 \\ b^{2} - 2b - 3 = 0\ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (a - 2)^{2} = 49 \\ (b - 1)^{2} = 4\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} a - 2 = \pm 7 \\ b - 1 = \pm 2 \\ \end{matrix} \right.\ \Longrightarrow\]