ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 328

\[\boxed{\text{329.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{x^{2} - 5x + 3}{x - 5} - \frac{x^{2} + 5x + 1}{x + 5} =\]

\[= \frac{1}{4}\]

\[x + \frac{3}{x - 5} - \left( x + \frac{1}{x + 5} \right) = \frac{1}{4}\]

\[x + \frac{3}{x - 5} - x - \frac{1}{x + 5} = \frac{1}{4}\]

\[\frac{3 \cdot (x + 5) - (x - 5)}{(x - 5)(x + 5)} = \frac{1}{4}\]

\[\frac{3x + 15 - x + 5}{x^{2} - 25} = \frac{1}{4}\]

\[4 \cdot (2x + 20) = x^{2} - 25\]

\[8x + 80 = x^{2} - 25\]

\[x^{2} - 8x - 105 = 0\]

\[D = 16 + 105 = 121\]

\[x_{1,2} = 4 \pm 11 = - 7;15.\]

\[\textbf{б)}\ \frac{x^{2} + 6x + 10}{x + 3} - \frac{x^{2} - 6x + 7}{x - 3} =\]

\[= 7\frac{1}{8}\]

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

\[= \frac{57}{8}\]

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

\[= \frac{57}{8}\]

\[6 + \frac{x - 3 + 2 \cdot (x + 3)}{(x + 3)(x - 3)} = \frac{57}{8}\]

\[\frac{3x + 3}{x^{2} - 9} = \frac{57}{8} - 6\ \]

\[\frac{3x + 3}{x^{2} - 9} = \frac{9}{8},\ \ \frac{x + 1}{x^{2} - 9} = \frac{3}{8},\ \ \]

\[3 \cdot \left( x^{2} - 9 \right) = 8 \cdot (x + 1)\]

\[D = 16 + 3 \cdot 35 = 121\]

\[x_{1,2} = \frac{4 \pm 11}{3} = - \frac{7}{3};5.\]

\[Ответ:\ \ а) - 7;15;\ \ б) - \frac{7}{3};\ \ 5.\]