Решебник по алгебре 8 класс Мерзляк Задание 179

\[\boxed{\text{179\ (179).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\frac{b + 4}{b^{2} - 6b + 9}\ :\frac{b^{2} - 16}{2b - 6} -\]

\[- \frac{2}{b - 4} =\]

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

\[- \frac{2}{b - 4} =\]

\[= \frac{2}{(b - 4)(b - 3)} - \frac{2^{\backslash b - 3}}{b - 4} =\]

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

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

\[= \frac{- 2}{b - 3} = \frac{2}{3 - b}\]

\[2)\ \left( \frac{m - 1^{\backslash m - 1}}{m + 1} - \frac{m + 1^{\backslash m + 1}}{m - 1} \right)\ :\]

\[:\frac{4m}{m^{2} - 1} =\]

\[= \frac{m^{2} - 2m + 1 - m^{2} - 2m - 1}{(m + 1)(m - 1)} \cdot\]

\[\cdot \frac{m^{2} - 1}{4m} = \frac{- 4m\left( m^{2} - 1 \right)}{\left( m^{2} - 1 \right) \cdot 4m} = - 1\]

\[3)\ \frac{2x}{x^{2} - y^{2}}\ :\]

\[:\left( \frac{1}{x^{2} + 2xy + y^{2}} - \frac{1}{y^{2} - x^{2}} \right) =\]

\[= \frac{2x}{x^{2} - y^{2}}\ :\]

\[:\left( \frac{1^{\backslash x - y}}{(x + y)^{2}} - \frac{1^{\backslash x + y}}{(x + y)(y - x)} \right) =\]

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

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

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

\[4)\ \left( \frac{2a - 3}{a^{2} - 4a + 4} - \frac{a - 1}{a^{2} - 2a} \right)\ :\]

\[:\frac{a^{2} - 2}{a^{3} - 4a} =\]

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

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

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

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

\[= \frac{\left( a^{2} - 2 \right)a(a - 2)(a + 2)}{a(a - 2)^{2}\left( a^{2} - 2 \right)} =\]

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