ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 951

\[1)\ \left\{ \begin{matrix} \sqrt{x + y - 1} = 1\ \ \ \ \ \ \ \ \ \ \\ \sqrt{x - y + 2} = 2y - 2 \\ \end{matrix} \right.\ \]

\[\sqrt{x + y - 1} = 1\]

\[x + y - 1 = 1\]

\[y = 2 - x.\]

\[\sqrt{x - y + 2} = 2y - 2\]

\[\sqrt{x - (2 - x) + 2} = 2(2 - x) - 2\]

\[\sqrt{2x} = 4 - 2x - 2\]

\[\sqrt{2x} = 2 - 2x\]

\[2x = 4 - 8x + 4x^{2}\]

\[4x^{2} - 10x + 4 = 0\]

\[2x^{2} - 5x + 2 = 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{5 - 3}{2 \bullet 2} = 0,5;\]

\[x_{2} = \frac{5 + 3}{2 \bullet 2} = 2;\]

\[y_{1} = 2 - 0,5 = 1,5;\]

\[y_{2} = 2 - 2 = 0.\]

\[Область\ определения:\]

\[2y - 2 > 0\]

\[2y > 2\]

\[y > 1.\]

\[Ответ:\ \ (0,5;\ 1,5).\]

\[2)\ \left\{ \begin{matrix} \sqrt{3y + x + 1} = 2\ \ \ \ \ \ \ \ \ \ \\ \sqrt{2x - y + 2} = 7y - 6 \\ \end{matrix} \right.\ \]

\[\sqrt{3y + x + 1} = 2\]

\[3y + x + 1 = 4\]

\[x = 3 - 3y.\]

\[\sqrt{2x - y + 2} = 7y - 6\]

\[\sqrt{2(3 - 3y) - y + 2} = 7y - 6\]

\[\sqrt{8 - 7y} = 7y - 6\]

\[8 - 7y = 49y^{2} - 84y + 36\]

\[49y^{2} - 77y + 28 = 0\]

\[D = 5929 - 5488 = 441\]

\[y_{1} = \frac{77 - 21}{2 \bullet 49} = \frac{4}{7};\]

\[y_{2} = \frac{77 + 21}{2 \bullet 49} = 1;\]

\[x_{1} = 3 - \frac{12}{7} = \frac{9}{7};\]

\[x_{2} = 3 - 3 = 0.\]

\[Область\ определения:\]

\[7y - 6 > 0\]

\[7y > 6\]

\[y > \frac{6}{7}.\]

\[Ответ:\ \ (0;\ 1).\]