ГДЗ по алгебре 8 класс Макарычев Задание 693

\[\boxed{\text{693\ (693).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 2x + 3 = \frac{34}{x - 5}\ \ \ \ \ \ | \cdot (x - 5)\]

\[x - 5 \neq 0,\ \ x \neq 5\]

\[(2x + 3)(x - 5) = 34\]

\[2x^{2} - 10x + 3x - 15 = 34\]

\[D = 49 + 392 = 441 = 21^{2}\]

\[x_{1,2} = \frac{7 \pm 21}{4} = 7;\ - 3,5\]

\[y_{1} = 2x + 3 = 14 + 3 = 17\]

\[y_{2} = - 7 + 3 = - 4\]

\[Ответ:(7;17)\ и\ ( - 3,5;\ - 4).\]

\[\textbf{б)}\frac{x^{2} - 5x}{x + 3} = 2x\ \ \ \ \ \ | \cdot (x + 3)\]

\[x + 3 \neq 0,\ \ x \neq - 3\]

\[x^{2} - 5x = 2x^{2} + 6x\]

\[2x^{2} - x^{2} + 6x + 5x = 0\]

\[x^{2} + 11x = 0\]

\[x(x + 11) = 0\]

\[Ответ:(0;0)\ и\ ( - 11; - 22)\text{.\ }\]