Решебник по алгебре 8 класс Мерзляк Задание 848

\[\boxed{\mathbf{848\ (848).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{7y}{y^{2} - 4} - \frac{14}{y^{2} - 4} =\]

\[= \frac{7 \cdot (y - 2)}{(y - 2)(y + 2)} = \frac{7}{y + 2}\]

\[2)\ \frac{y^{2} - 3y}{25 - y^{2}} - \frac{7y - 25}{25 - y^{2}} =\]

\[= \frac{y^{2} - 3y - 7y + 25}{(5 - y)(5 + y)} =\]

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

\[3)\ \frac{9p + 5}{3p + 6} - \frac{10p - 12}{3p + 6} + \frac{9p - 1}{3p + 6} =\]

\[= \frac{9p + 5 - 10p + 12 + 9p - 1}{3p + 6} =\]

\[= \frac{8p + 16}{3p + 6} = \frac{8 \cdot (p + 2)}{3 \cdot (p + 2)} = \frac{8}{3}\]

\[4)\ \frac{7x + 5}{3 - x} + \frac{5x + 11}{x - 3} =\]

\[= \frac{7x + 5 - 5x - 11}{3 - x} = \frac{2x - 6}{3 - x} =\]

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

\[5)\ \frac{(3a - 1)^{2}}{4a - 4} + \frac{(a - 3)^{2}}{4 - 4a} =\]

\[= \frac{9a^{2} - 6a + 1 - a^{2} + 6a - 9}{4a - 4} =\]

\[= \frac{8a^{2} - 8}{4 \cdot (a - 1)} =\]

\[= \frac{8 \cdot (a - 1)(a + 1)}{4 \cdot (a - 1)} = 2a + 2\]

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

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

\[= 1\]

\[7)\ \frac{7}{a - 2} - \frac{b}{2 - a} = \frac{7 + b}{a - 2}\]

\[8)\ \frac{6a}{5 - a} - \frac{4a}{a - 5} = \frac{6a + 4a}{5 - a} =\]

\[= \frac{10a}{5 - a}\]