ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 830

\[1)\ 2x^{2}y^{2} - 14y^{2} = 25 - x^{2}\]

\[\left( 2x^{2} - 14 \right)y^{2} = 25 - x^{2}\]

\[y^{2} = \frac{25 - x^{2}}{2x^{2} - 14}\]

\[y = \pm \sqrt{\frac{25 - x^{2}}{2x^{2} - 14}}.\]

\[Область\ определения:\]

\[\frac{25 - x^{2}}{2x^{2} - 14} \geq 0\]

\[\frac{x^{2} - 25}{x^{2} - 7} \leq 0\]

\[\frac{(x + 5)(x - 5)}{\left( x + \sqrt{7} \right)\left( x - \sqrt{7} \right)} \leq 0\]

\[- 5 \leq x < - \sqrt{7}\text{\ \ \ }\]

\[\sqrt{7} < x \leq 5.\]

\[x = \pm 5:\]

\[y = \pm \sqrt{\frac{25 - 25}{100 - 14}} = \pm \sqrt{\frac{0}{86}} = 0.\]

\[x = \pm 4:\]

\[y = \pm \sqrt{\frac{25 - 16}{32 - 14}} = \pm \sqrt{\frac{9}{18}} = \pm \frac{1}{\sqrt{9}}.\]

\[x = \pm 3:\]

\[y = \pm \sqrt{\frac{25 - 9}{18 - 14}} = \pm \sqrt{\frac{16}{4}} = \pm 2.\]

\[Ответ:\ \ \]

\[( - 5;\ 0);\ ( - 3;\ - 2);\ ( - 3;\ 2);\ \]

\[(3;\ - 2);\ (3;\ 2);\ (5;\ 0).\]

\[2)\ 3x^{2} - 8xy - 16y^{2} = 19\]

\[D = (8x)^{2} + 4 \bullet 3 \bullet 16y^{2} =\]

\[= 64y^{2} + 192y^{2} = 256y^{2}\]

\[x_{1} = \frac{8y - 16y}{2 \bullet 3} = - \frac{8y}{6} = - \frac{4y}{3};\]

\[x_{2} = \frac{8y + 16y}{2 \bullet 3} = \frac{24y}{6} = 4y;\]

\[3\left( x + \frac{4y}{3} \right)(x - 4y) = 19\]

\[(3x + 4y)(x - 4y) - 19 = 0.\]

\[Первая\ пара\ значений:\]

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

\[4y = \pm 1 - 3x\]

\[4x = \pm 20\]

\[x = \pm 5.\]

\[x - 4y = \pm 19\]

\[x \mp 1 + 3x = \pm 19\text{\ \ }\]

\[y = \frac{\pm 1 \mp 15}{4} = \mp \frac{14}{4} = - 3,5.\]

\[Вторая\ пара\ значений:\]

\[3x + 4y = \pm 19\text{\ \ }\]

\[4y = \pm 19 - 3x\ \ \ \]

\[4x = \pm 20\text{\ \ }\]

\[x = \pm 5.\]

\[x - 4y = \pm 1\]

\[x \mp 19 + 3x = \pm 1\]

\[y = \frac{\pm 19 \mp 15}{4} = \pm \frac{4}{4} = \pm 1.\]

\[Ответ:\ \ ( - 5;\ - 1);\ (5;\ 1).\ \]