ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 238

\[\boxed{\text{239.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

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

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

\[\frac{1}{x - 4} - \frac{1}{x - 5} = \frac{1}{x + 4} - \frac{1}{x - 2}\]

\[ОДЗ:\ \ x \neq 4;5;\ - 4;2.\]

\[- (x - 2)(x + 4) =\]

\[= - 6 \cdot (x - 4)(x - 5);\]

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

\[= - 6 \cdot \left( x^{2} - 9x + 20 \right);\]

\[5x^{2} - 56x + 128 = 0\]

\[D = 28^{2} - 5 \cdot 128 = 144\]

\[x_{1,2} = \frac{28 \pm 12}{5} = 8;3,2.\]

\[Ответ:x = 8;\ \ x = 3,2.\]

\[\textbf{б)}\ \frac{1}{x + 1} + \frac{1}{x + 3} = \frac{1}{x + 28} + \frac{1}{x};\]

\[\frac{1}{x + 1} - \frac{1}{x - 28} = \frac{1}{x} - \frac{1}{x + 3};\]

\[ОДЗ:\ \ \ x \neq - 1;28;0;\ - 3.\]

\[- 29x(x + 3) =\]

\[= 3 \cdot (x + 1)(x - 28);\]

\[8x^{2} - 2x - 28 = 0\]

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

\[D = 1 + 4 \cdot 4 \cdot 14 = 225\]

\[x_{1,2} = \frac{1 \pm 15}{8} = 2;\ - 1,75.\]

\[Ответ:x = 2;\ \ x = - 1,75.\]