Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 392

\[\boxed{\mathbf{392}.}\]

\[1)\ \left\{ \begin{matrix} x - xy = 0\ \ \ \\ y^{2} + 3xy = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x(1 - y) = 0 \\ y^{2} + 3xy = 4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 0\ \ \ \\ y^{2} = 4\ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = 0\ \ \ \\ y = \pm 2 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 1 - y = 0\ \ \ \ \ \ \ \\ y^{2} + 3xy = 4 \\ \end{matrix} \right.\ \ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 1\ \ \\ 3x = 3 \\ \end{matrix} \right.\ \ \Longrightarrow \left\{ \begin{matrix} y = 1 \\ x = 1 \\ \end{matrix} \right.\ \]

\[Ответ:(0;\ - 2);(0;2);(1;1).\]

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

\[y^{2} + y = 5y\]

\[y^{2} - 4y = 0\]

\[y(y - 4) = 0\]

\[y = 0;\ \ \ y = 4.\]

\[\ \left\{ \begin{matrix} y = 0\ \ \ \\ x^{2} = y \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y = 0 \\ x = 0 \\ \end{matrix} \right.\ ;\]

\(\left\{ \begin{matrix} y = 4\ \\ x^{2} = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 4\ \ \ \\ x = \pm 2 \\ \end{matrix} \right.\ \)

\[Ответ:(0;0);(2;4);(2;\ - 4).\]

\[3)\ \left\{ \begin{matrix} xy + x - 3y = 3 \\ x^{2} + y^{2} = 10\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[xy + x - 3y = 3\]

\[x(y + 1) = 3y + 3\]

\[x = \frac{3y + 3}{y + 1}.\]

\[\left\{ \begin{matrix} x = \frac{3y + 3}{y + 1}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{(3 + 3y)^{2}}{(y + 1)^{2}} + y^{2} = 10 \\ \end{matrix} \right.\ \]

\[9 + 18y + 9y^{2} + y^{2}(y + 1)^{2} =\]

\[= 10 \cdot (y + 1)^{2}\]

\[9 + 18y + 9y^{2} +\]

\[+ y^{2}\left( y^{2} - 2y + 1 \right) =\]

\[= 10 \cdot \left( y^{2} + 2y + 1 \right)\]

\[y^{4} + 2y^{3} + y^{2} + 9 + 18y +\]

\[+ 9y^{2} - 10y^{2} - 20y - 10 = 0\]

\[y^{4} + 2y^{3} - 2y - 1 = 0\]

\[(y - 1)(y + 1)\left( y^{2} + 2y + 1 \right) = 0\]

\[(y - 1)(y + 1)(y + 1)^{2} = 0\]

\[(y - 1)(y + 1)^{3} = 0\]

\[y = \pm 1:\]

\[y^{2} = 1.\]

\[x^{2} = 10 - 1 = 9\]

\[x = \pm 3.\]

\[Ответ:(3;1);(3;\ - 1);\]

\[( - 3;1);( - 3;\ - 1).\]

\[4)\ \left\{ \begin{matrix} x + y + xy = 11\ \\ x^{2}y + xy^{2} = 30 \\ \end{matrix} \right.\ \]

\[x + y + xy = 11\]

\[x(1 + y) = 11 - y\]

\[x = \frac{11 - y}{1 + y}.\]

\[Подставим\ во\ второе\]

\[\ уравнение:\]

\[\left( \frac{11 - y}{1 + y} \right)^{2} \cdot y + \frac{11 - y}{1 + y} \cdot y^{2} = 30\]

\[\left( 121 - 22y + y^{2} \right)y +\]

\[+ \left( 11y^{2} - y^{3} \right)(1 + y) =\]

\[= 30 \cdot (1 + y)^{2}\ \]

\[121y - 22y^{2} + y^{3} + 11y^{2} -\]

\[- y^{3} + 11y^{3} - y^{4} = 30 +\]

\[+ 60y + 30y^{2}\]

\[y^{4} - 11y^{3} + 41y^{2} -\]

\[- 61y + 30 = 0\]

\[Делители:\ \pm 1;\ \pm 2;\ \pm 3;\ \pm 5;\ \]

\[\pm 6;\ \pm 10;\ \pm 15;\ \pm 30.\]

\[P(y) = (y - 1)(y - 2)(y - 3)(y - 5) = 0.\]

\[y = 1 \Longrightarrow x = 5;\]

\[y = 2 \Longrightarrow x = 3;\]

\[y = 3 \Longrightarrow x = 2;\]

\[y = 5 \Longrightarrow x = 1.\]

\[Ответ:(5;1);(3;2);(2;3);(1;5).\]