Решебник по алгебре 8 класс Мерзляк ФГОС Задание 1065

\[\boxed{\mathbf{106}\mathbf{5}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{x^{5}y^{7} - x^{3}y^{9}}{x^{3}y^{7}} =\]

\[= \frac{x^{3}y^{7}\left( x^{2} - y^{2} \right)}{x^{3}y^{7}} =\]

\[= (x - y)(x + y)\]

\[при\ \ \ \ x = - 0,2;\ \ \ y = 0,5:\]

\[( - 0,2 - 0,5)( - 0,2 + 0,5) =\]

\[= - 0,7 \cdot 0,3 = - 0,21\]

\[2)\ \frac{4a^{2} - 36}{5a^{2} - 30a + 45} =\]

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

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

\[при\ \ \ a = 2:\]

\[\frac{4 \cdot 5}{5 \cdot ( - 1)} = - 4\]

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

\[= \frac{9 \cdot (a + b)^{2}}{3 \cdot (a + b)(a - b)} =\]

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

\[при\ \ \ a = \frac{1}{3},\ b = - \frac{1}{6}:\]

\[\frac{3 \cdot \left( \frac{2}{6} - \frac{1}{6} \right)}{\left( \frac{2}{6} + \frac{1}{6} \right)} = \frac{1}{2}\ :\frac{1}{2} = 1\]

\[4)\ \frac{20x^{2} - 140xy + 245y^{2}}{4x - 14y} =\]

\[= \frac{5 \cdot (2x - 7y)^{2}}{2 \cdot (2x - 7y)} = \frac{5 \cdot (2x - 7y)}{2}\]

\[при\ 2x - 7y = - 0,5,\]

\[2x = 7y - 0,5:\]

\[\frac{5 \cdot (7y - 0,5 - 7y)}{2} = - 1,25\]