ГДЗ по алгебре 9 класс Макарычев Задание 293

\[\boxed{\text{293}\text{\ (293)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[ОДЗ:\ \ x \neq 7;1;10;9.\]

\[(x - 1)(x - 10)(x - 9) -\]

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

\[(x - 7)(x - 1)(x - 9) -\]

\[- (x - 7)(x - 1)(x - 10)\]

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

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

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

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

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

\[D_{1} = 53^{2} - 5 \cdot 533 = 144\]

\[x_{1} = \frac{53 + 12}{5} = 13;\ \ x_{2} =\]

\[= \frac{53 - 12}{5} = 8,2.\]

\[Ответ:x = 13;\ \ x = 8,2.\]

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

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

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

\[(x + 9)(x + 5)(x + 21) -\]

\[- (x + 3)(x + 5)(x + 21) =\]

\[= (x + 3)(x + 9)(x + 21) -\]

\[- (x + 3)(x + 9)(x + 5)\]

\[(x + 5)(x + 21)(x + 9 - x - 3) =\]

\[= (x + 3)(x + 9)(x + 21 - x - 5)\]

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

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

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

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

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

\[D_{1} = 9^{2} + 5 \cdot 99 = 576\]

\[x_{1} = \frac{- 9 + 24}{5} = 3;\ \ \]

\[x_{2} = \frac{- 9 - 24}{5} = - 6,6.\]

\[Ответ:x = 3;\ \ x = - 6,6.\]