Решебник по алгебре 9 класс Макарычев Задание 371

 

\[\boxed{\text{371\ (371).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{x^{4}\ }{x^{2} - 2} + \frac{1 - 4x^{2}}{2 - x^{2}} + 4 = 0\]

\[\frac{x^{4}}{x^{2} - 2} - \frac{1 - 4x^{2}}{x^{2} - 2} + 4^{\backslash x^{2} - 2} = 0\]

\[\frac{x^{4} - 1 + 4x^{2} + 4x^{2} - 8}{x^{2} - 2} = 0\]

\[ОДЗ:x^{2} - 2 \neq 0,\ \ x^{2} \neq 2,\]

\[\ \ x \neq \pm \sqrt{2}.\]

\[x^{4} + 8x^{2} - 9 = 0\]

\[Пусть\ \ y = x^{2};\ \ \ \ y \geq 0:\]

\[y^{2} + 8y - 9 = 0\ \ \]

\[D_{1} = 16 + 9 = 25\]

\[y_{1} = - 4 + 5 = 1;\ \ y_{2} =\]

\[= - 4 - 5 = - 9.\]

\[Так\ как\ y \geq 0,\ то\ y = 1:\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[Ответ:\ \ x = \pm 1.\]

\[\textbf{б)}\ \frac{x^{2} + 3}{x^{2} + 1} + \frac{2}{x^{2} - 4} +\]

\[+ \frac{10}{x^{4} - 3x^{2} - 4} = 0\]

\[Пусть\ \ y = x^{2};\ \ y \geq 0:\]

\[\frac{y + 3}{y + 1} + \frac{2}{y - 4} + \frac{10}{y^{2} - 3y - 4} = 0\]

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

\[y_{1} + y_{2} = 3;\ \ y_{1} \cdot y_{2} = - 4\]

\[y_{1} = 4;\ \ y_{2} = - 1.\]

\[\frac{y + 3^{\backslash y - 4}}{y + 1} + \frac{2^{\backslash y + 1}}{y - 4} +\]

\[+ \frac{10}{(y - 4)(y + 1)} = 0\]

\[\frac{(y + 3)(y - 4) + 2 \cdot (y + 1) + 10}{(y - 4)(y + 1)} = 0\]

\[ОДЗ:\ \ \ y \neq 4;\ - 1.\]

\[y^{2} - 4y + 3y - 12 + 2y +\]

\[+ 2 + 10 = 0\]

\[y^{2} + y = 0\]

\[y(y + 1) = 0\]

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

\[y \geq 0:\]

\[x^{2} = 0\]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]

\[\boxed{\text{371.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

Пояснение.

Решение.

\[(x - 4)^{2} + (y + m)^{2} = 15\]

\[a = 4;\ \ b = - m.\]

\[Центр\ окружности\ расположен\ \]

\[в\ точке\ O(4; - m).\]

\[Так\ как\ центр\ располагается\ в\ \]

\[ІV\ координатной\ четверти:\]

\[m < 0.\]

\[Ответ:при\ m < 0.\]