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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 361
Задание 361
\[\boxed{\mathbf{361}.}\]
\[1)\ \left\{ \begin{matrix} y - x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{3} - 4xy + 5y = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} y = x + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{3} - 4x(x + 1) + 5 \cdot (x + 1) = - 1 \\ \end{matrix} \right.\ \]
\[x^{3} - 4x^{2} - 4x + 5x + 5 + 1 = 0\]
\[x^{3} - 4x^{2} + x + 6 = 0\]
| \[1\] | \[- 4\] | \[1\] | \[6\] | |
|---|---|---|---|---|
| \[- 1\] | \[1\] | \[- 5\] | \[6\] | \[0\] |
| \[2\] | \[1\] | \[- 3\] | \[0\] |
\[(x + 1)(x - 2)(x - 3) = 0\]
\[x_{1} = - 1;\ \ \ y_{1} = - 1 + 1 = 0;\]
\[x_{2} = 2;\ \ \ \ \ \ y_{2} = 2 + 1 = 3;\]
\[x_{3} = 3;\ \ \ \ \ \ y_{3} = 3 + 1 = 4.\]
\[Ответ:( - 1;0);(2;3);(3;4).\]
\[2)\ \left\{ \begin{matrix} x - y = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{3} + 9xy + 25y + 44 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\left\{ \begin{matrix} y = x - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{3} + 9x(x - 2) + 25 \cdot (x - 2) + 44 = 0 \\ \end{matrix} \right.\ \]
\[2x^{3} + 9x^{2} - 18x + 25x -\]
\[- 50 + 44 = 0\]
\[2x^{3} + 9x^{2} + 7x - 6 = 0\]
\[(x + 2)(x + 3)(2x - 1) = 0\]
| \[2\] | \[9\] | \[7\] | \[- 6\] | |
|---|---|---|---|---|
| \[- 2\] | \[2\] | \[5\] | \[- 3\] | \[0\] |
| \[- 3\] | \[2\] | \[- 1\] | \[0\] |
\[x = - 2:\]
\[y = - 2 - 2 = - 4.\]
\[x = - 3:\]
\[y = - 2 - 3 = - 5.\]
\[x = 0,5:\]
\[y = 0,5 - 2 = - 1,5.\]
\[Ответ:( - 2; - 4);( - 3;\ - 5);\]
\[(0,5; - 1,5).\]
\[3)\ \left\{ \begin{matrix} x - y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 6x^{2}y + xy - y = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = x - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 6x^{2}(x - 1) + x(x - 1) - (x - 1) = 0 \\ \end{matrix} \right.\ \]
\[(x - 1)\left( 6x^{2} + x - 1 \right) = 0\]
\[6x^{2} + x - 1 = 0\]
\[D = 1 + 24 = 25\]
\[x_{1} = \frac{- 1 + 5}{12} = \frac{1}{3};\ \ \]
\[\ x_{2} = \frac{- 1 - 5}{12} = - \frac{1}{2} = - 0,5.\]
\[x = 1:\]
\[y = 1 - 1 = 0.\]
\[x = \frac{1}{3}:\]
\[y = \frac{1}{3} - 1 = - \frac{2}{3}.\]
\[x = - 0,5:\]
\[y = - 0,5 - 1 = - 1,5.\]
\[Ответ:\left( \frac{1}{3}; - \frac{2}{3} \right);( - 0,5;\ - 1,5);\]
\[(1;0).\]
\[4)\ \left\{ \begin{matrix} y - x = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{3}y + 9x^{2}y - 5xy = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} y = x + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{3}(x + 2) + 9x^{2}(x + 2) - 5x(x + 2) = 0 \\ \end{matrix} \right.\ \]
\[(x + 2)\left( 2x^{3} + 9x^{2} - 5x \right) = 0\]
\[x(x + 2)\left( 2x^{2} + 9x - 5 \right) = 0\]
\[2x^{2} + 9x - 5 = 0\]
\[D = 81 + 40 = 121\]
\[x_{1} = \frac{- 9 - 11}{4} = - 5;\ \ \ \]
\[x_{2} = \frac{- 9 + 11}{4} = \frac{2}{4} = 0,5.\]
\[x = 0:\]
\[y = 0 + 2 = 2.\]
\[x = - 2:\]
\[y = - 2 + 2 = 0.\]
\[x = - 5:\]
\[y = - 5 + 2 = - 3.\]
\[x = 0,5:\]
\[y = 0,5 + 2 = 2,5.\]
\[Ответ:(0,5;2,5);( - 2;0);(0;2);\]
\[( - 5; - 3).\]
