ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 237

\[\boxed{\text{238.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{1}{x - 7} - \frac{1}{x - 1} =\]

\[= \frac{1}{x - 10} - \frac{1}{x - 9}\]

\[ОДЗ:\ \ x \neq 7;1;10;9.\]

\[(x - 9)(x - 10)(x - 1 - x + 7) =\]

\[= (x - 1)(x - 7)(x - 9 - x + 10);\]

\[(x - 9)(x - 10) \cdot 6 =\]

\[= (x - 1)(x - 7) \cdot 1;\]

\[5x^{2} - 106x + 533 = 0\]

\[D = 53^{2} - 5 \cdot 533 = 144\]

\[x_{1,2} = \frac{53 \pm 12}{5},\ \ \]

\[x_{1} = 13,\ \ x_{2} = 8,2.\]

\[Ответ:x = 13;\ \ x = 8,2.\]

\[\textbf{б)}\ \frac{1}{x + 3} - \frac{1}{x + 9} =\]

\[= \frac{1}{x + 5} - \frac{1}{x + 21}\]

\[ОДЗ:\ \ x \neq - 3;\ - 9;\ - 5;\ - 21.\]

\[6 \cdot (x + 5)(x + 21) =\]

\[= 16 \cdot (x + 3)(x + 9);\]

\[6 \cdot \left( x^{2} + 21x + 5x + 105 \right) =\]

\[= 16 \cdot \left( x^{2} + 9x + 3x + 27 \right);\]

\[5x^{2} + 18x - 99 = 0\]

\[D = 9^{2} + 5 \cdot 99 = 576\]

\[x_{1,2} = \frac{- 9 \pm 24}{5},\ \ \]

\[x_{1} = 3,\ \ x_{2} = - 6,6.\]

\[Ответ:x = 3;\ \ x = - 6,6.\]