ГДЗ по алгебре 8 класс Макарычев Задание 234

\[\boxed{\text{234\ (234).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{x^{2} - 3x + 6}{x - 3} =\]

\[= \frac{x^{2} - 3x}{x - 3} + \frac{6}{x - 3} =\]

\[= \frac{x(x - 3)}{x - 3} + \frac{6}{x - 3} = x + \frac{6}{x - 3}\]

\[\textbf{б)}\ \frac{y^{2} + 5y - 8}{y + 5} =\]

\[\frac{y^{2} + 5y}{y + 5} - \frac{8}{y + 5} =\]

\[= \frac{y(y + 5)}{y + 5} - \frac{8}{y + 5} = y - \frac{8}{y + 5}\]

\[\textbf{в)}\ \frac{a^{2} + 7a + 2}{a + 6} =\]

\[= \frac{a^{2} + 6a + a + 2}{a + 6} =\]

\[= \frac{a^{2} + 6a}{a + 6} + \frac{a + 2}{a + 6} =\]

\[= \frac{a(a + 6)}{a + 6} + \frac{a + 2}{a + 6} =\]

\[= a + \frac{a + 2}{a + 6}\]

\[\textbf{г)}\ \frac{3b^{2} - 10b - 1}{b - 3} =\]

\[= \frac{3b^{2} - 9b - b - 1}{b - 3} =\]

\[= \frac{3b^{2} - 9b}{b - 3} - \frac{b + 1}{b - 3} =\]

\[= \frac{3b(b - 3)}{b - 3} - \frac{b + 1}{b - 3} =\]

\[= 3b - \frac{b + 1}{b - 3}\]