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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1429
Задание 1429
\[\boxed{\mathbf{1429}\mathbf{.}}\]
\[1)\ \left\{ \begin{matrix} \sqrt{x + y - 1} = 1\ \ \ \ \ \ \ \ \ \ \\ \sqrt{x - y + 2} = 2y - 2 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x + y - 1 = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x - y + 2 = 4y^{2} - 8y + 4 \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x = 2 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} - 7y - x + 2 = 0 \\ \end{matrix} \right.\ \]
\[4y^{2} - 7y - (2 - y) + 2 = 0\]
\[4y^{2} - 6y = 0\]
\[2y \bullet (2y - 3) = 0\]
\[y_{1} = 0;\ \ \ y_{2} = 1,5;\]
\[x_{1} = 2 - 0 = 2;\]
\[x_{2} = 2 - 1,5 = 0,5.\]
\[Имеет\ смысл\ при:\]
\[\left\{ \begin{matrix} x + y - 1 \geq 0 \\ x - y + 2 \geq 0 \\ \end{matrix} \right.\ \ ( - )\]
\[2y - 3 \geq 0\]
\[y \geq \frac{3}{2}.\]
\[\left\{ \begin{matrix} x + y - 1 \geq 0 \\ x - y + 2 \geq 0 \\ \end{matrix} \right.\ \ ( + )\]
\[2x - 1 \geq 0\]
\[x \geq \frac{1}{2}.\]
\[2y - 2 \geq 0\]
\[y - 1 \geq 0\]
\[y \geq 1.\]
\[Ответ:\ \ (0,5;\ 1,5).\]
\[2)\ \left\{ \begin{matrix} \sqrt{3y + x + 1} = 2\ \ \ \ \ \ \ \ \ \ \\ \sqrt{2x - y + 2} = 7y - 6 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 3y + x + 1 = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x - y + 2 = 49y^{2} - 84y + 36 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 3 - 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 49y^{2} - 83y - 2x + 34 = 0 \\ \end{matrix} \right.\ \]
\[49y^{2} - 83y - 2(3 - 3y) + 34 = 0\]
\[49y^{2} - 83y - 6 + 6y + 34 = 0\]
\[49y^{2} - 77y + 28 = 0\]
\[D = 5929 - 5488 = 441\]
\[y_{1} = \frac{77 - 21}{2 \bullet 49} = \frac{56}{98} = \frac{4}{7};\ \]
\[y_{2} = \frac{77 + 21}{2 \bullet 49} = \frac{98}{98} = 1;\]
\[x_{1} = 3 - 3 \bullet \frac{4}{7} = \frac{21}{7} - \frac{12}{7} = \frac{9}{7};\]
\[x_{2} = 3 - 3 \bullet 1 = 0.\]
\[Имеет\ смысл\ при:\]
\[\left\{ \begin{matrix} 3y + x + 1 \geq 0\ \ \ \ \ | \bullet 2 \\ 2x - y + 2 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} 2x + 6y + 2 \geq 0 \\ 2x - y + 2 \geq 0\ \ \\ \end{matrix} \right.\ \ ( - )\]
\[7y \geq 0\]
\[y \geq 0.\]
\[\left\{ \begin{matrix} 3y + x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ 2x - y + 2 \geq 0\ \ \ \ \ | \bullet 3 \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} 3y + x + 1 \geq 0\ \ \ \ \ \ \\ - 3y + 6x + 6 \geq 0 \\ \end{matrix} \right.\ \ ( + )\]
\[7x + 7 \geq 0\]
\[x + 1 \geq 0\]
\[x \geq - 1.\]
\[7y - 6 \geq 0\]
\[7y \geq 6\]
\[y \geq \frac{6}{7}.\]
\[Ответ:\ \ (0;\ 1).\]
