Решебник по алгебре 9 класс Макарычев Задание 368

 

\[\boxed{\text{368\ (368).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\frac{1}{x^{3} - x^{2} + x - 1} +\]

\[+ \frac{4x^{2} + 21}{x^{3} + x^{2} + x + 1} =\]

\[= \frac{4x^{3} - 3x^{2} + 14x - 4}{x^{4} - 1}\]

\[\frac{1}{x^{2}(x - 1) + (x - 1)} +\]

\[+ \frac{4x^{2} + 21}{x^{2}(x + 1) + (x + 1)} =\]

\[= \frac{4x^{3} - 3x^{2} + 14x - 4}{\left( x^{2} - 1 \right)\left( x^{2} + 1 \right)}\]

\[\frac{1^{\backslash x + 1}}{(x - 1)\left( x^{2} + 1 \right)} +\]

\[+ \frac{4x^{2} + 21^{\backslash x - 1}}{(x + 1)\left( x^{2} + 1 \right)} =\]

\[= \frac{4x^{3} - 3x^{2} + 14x - 4}{\left( x^{2} - 1 \right)\left( x^{2} + 1 \right)}\]

\[ОДЗ:\ \ \ x \neq \pm 1.\]

\[x + 1 + 4x^{3} - 4x^{2} + 21x - 21 -\]

\[- 4x^{3} + 3x^{2} - 14x + 4 = 0\]

\[- x^{2} + 8x - 16 = 0\]

\[x^{2} - 8x + 16 = 0\]

\[(x - 4)^{2} = 0\]

\[x - 4 = 0\]

\[x = 4.\]

\[Ответ:x = 4.\]

\[\boxed{\text{368.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

Пояснение.

Решение.

\[\text{K\ }(2;\ - 5):\]

\[(x - 2)^{2} + (y + 5)^{2} = r^{2}.\]

\[\textbf{а)}\ A\ ( - 1; - 1):\]

\[( - 1 - 2)^{2} + ( - 1 + 5)^{2} =\]

\[= 9 + 16 = 25\]

\[(x - 2)^{2} + (y + 5)^{2} = 25.\]

\[\textbf{б)}\ B\ ( - 3;7):\]

\[( - 3 - 2)^{2} + (7 + 5)^{2} =\]

\[= 25 + 141 = 169\]

\[(x - 2)^{2} + (y + 5)^{2} = 169.\]

\[\textbf{в)}\ C\ (1;\ - 4):\]

\[(1 - 2)^{2} + ( - 4 + 5)^{2} =\]

\[= 1 + 1 = 2\]

\[(x - 2)^{2} + (y + 5)^{2} = 2.\]