ГДЗ по алгебре 8 класс Мерзляк Задание 850

\[\boxed{\mathbf{850\ (850).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{2a - 1}{a - 4} - \frac{3a + 2}{2 \cdot (a - 4)} =\]

\[= \frac{4a - 2 - 3a - 2}{2 \cdot (a - 4)} =\]

\[= \frac{(a - 4)}{2 \cdot (a - 4)} = \frac{1}{2}\]

\[2)\ \frac{x + 2}{3x + 9} - \frac{4 - x}{5x + 15} =\]

\[= \frac{x + 2}{3 \cdot (x + 3)} - \frac{4 - x}{5 \cdot (x + 3)} =\]

\[= \frac{5x + 10 - 12 + 3x}{15 \cdot (x + 3)} =\]

\[= \frac{8x - 2}{15 \cdot (x + 3)}\]

\[3)\ \frac{m + 1}{m - 3} - \frac{m + 2}{m + 3} =\]

\[= \frac{m^{2} + 4m + 3 - m^{2} + m + 6}{(m - 3)(m + 3)} =\]

\[= \frac{5m + 9}{m² - 9}\]

\[4)\ \frac{x}{x + y} - \frac{2y^{2}}{y^{2} - x^{2}} - \frac{y}{x - y} =\]

\[= \frac{x^{2} - xy + 2y^{2} - xy - y^{2}}{(x + y)(x - y)} =\]

\[= \frac{x^{2} - 2xy + y^{2}}{(x + y)(x - y)} =\]

\[= \frac{(x - y)²}{(x + y)(x - y)} = \frac{x - y}{x + y}\]

\[5)\ \frac{m}{3m - 2n} - \frac{3m^{2} - 3mn}{9m^{2} - 12m + 4n^{2}} =\]

\[= \frac{3m^{2} - 2mn - 3m^{2} + 3mn}{(3m - 2n)^{2}} =\]

\[= \frac{\text{mn}}{(3m - 2n)²}\]

\[6)\ \frac{a + 3}{a^{2} - 2a} - \frac{a - 2}{5a - 10} + \frac{a + 2}{5a} =\]

\[= \frac{a + 3}{a(a - 2)} - \frac{a - 2}{5 \cdot (a - 2)} + \frac{a + 2}{5a} =\]

\[= \frac{5a + 15 - a^{2} + 2a + a^{2} - 4}{5a(a - 2)} =\]

\[= \frac{7a + 11}{5a(a - 2)}\]

\[7)\ \frac{3}{3a - 3} - \frac{a - 1}{2a^{2} - 4a + 2} =\]

\[= \frac{3}{3 \cdot (a - 1)} - \frac{a - 1}{2 \cdot (a - 1)^{2}} =\]

\[= \frac{6a - 6 - 3a + 3}{6 \cdot (a - 1)^{2}} =\]

\[= \frac{3a - 3}{6 \cdot (a - 1)^{2}} = \frac{3 \cdot (a - 1)}{6 \cdot (a - 1)^{2}} =\]

\[= \frac{1}{2a - 2}\]

\[8)\ 2 - \frac{14}{m - 2} - m =\]

\[= \frac{2 \cdot (m - 2) - 14 - m(m - 2)}{m - 2} =\]

\[= \frac{2m - 4 - 14 - m^{2} + 2m}{m - 2} =\]

\[= \frac{- m^{2} + 4m - 18}{m - 2} =\]

\[= - \frac{m^{2} - 4m + 18}{m - 2}\]

\[= \frac{18x² + 90}{(x - 3)²(x + 3)²} = \frac{18x^{2} + 90}{\left( x^{2} - 9 \right)^{2}}\]