ГДЗ по алгебре 9 класс Макарычев Задание 294

\[\boxed{\text{294}\text{\ (294)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[ОДЗ:\ \ x \neq 4;5;\ - 4;2.\]

\[(x - 5)(x + 4)(x - 2) -\]

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

\[= (x - 4)(x - 5)(x - 2) -\]

\[- (x - 4)(x - 5)(x + 4)\]

\[(x - 2)(x + 4)(x - 5 - x + 4) =\]

\[= (x - 4)(x - 5)(x - 2 - x - 4)\]

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

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

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

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

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

\[= - 6x^{2} + 54x - 120\]

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

\[D_{1} = 28^{2} - 5 \cdot 128 = 144\]

\[x_{1} = \frac{28 + 12}{5} = 8;\ \ \ \]

\[x_{2} = \frac{28 - 12}{5} = 3,2.\]

\[Ответ:x = 8;\ \ x = 3,2.\]

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

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

\[ОДЗ:\ \ \ x \neq - 1;28;0;\ - 3.\]

\[(x + 28) \cdot x \cdot (x + 3) -\]

\[- x(x + 1)(x + 3) =\]

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

\[- x(x + 1)(x + 28)\]

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

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

\[27x(x + 3) = 3 \cdot (x + 1)(x + 28)\]

\[27x^{2} + 81x =\]

\[= 3 \cdot \left( x^{2} + 28x + x + 28 \right)\]

\[27x^{2} + 81x = 3x^{2} + 87x + 84\]

\[24x^{2} - 6x - 84 = 0\ \ \ \ |\ :6\]

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

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

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

\[Ответ:x = 2;\ \ x = - 1,75.\]