ГДЗ по алгебре 8 класс Макарычев Задание 32

\[\boxed{\text{32\ (32).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ a = - 2;\ b = - 0,1:\]

\[\frac{15a^{2} - 10ab}{3ab - 2b^{2}} = \frac{5a \cdot (3a - 2b)}{b \cdot (3a - 2b)} =\]

\[= \frac{5a}{b}\]

\[\frac{5 \cdot ( - 2)}{( - 0,1)} = \frac{- 10}{( - 0,1)} = 100.\]

\[\textbf{б)}\ c = \frac{2}{3};\ d = \frac{1}{2}:\]

\[\frac{9c^{2} - 4d^{2}}{18c^{2}d - 12cd^{2}} =\]

\[= \frac{(3c - 2d) \cdot (3c + 2d)}{6cd \cdot (3c - 2d)} =\]

\[= \frac{3c + 2d}{6cd}\]

\[\frac{3 \cdot \frac{2}{3} + 2 \cdot \frac{1}{2}}{6 \cdot \frac{2}{3} \cdot \frac{1}{2}} = \frac{2 + 1}{2 \cdot 1} = \frac{3}{2} = 1,5.\ \]

\[\textbf{в)}\ x = \frac{2}{3};\ y = - 0,4:\]

\[\frac{6x^{2} + 12xy}{5xy + 10y^{2}} = \frac{6x \cdot (x + 2y)}{5y \cdot (x + 2y)} =\]

\[= \frac{6x}{5y}\ \]

\[\frac{6 \cdot \frac{2}{3}}{5 \cdot ( - 0,4)} = \frac{4}{- 2} = - 2.\]

\[\textbf{г)}\ x = - 0,2;\ y = - 0,6:\]

\[\frac{x^{2} + 6xy + 9y^{2}}{4x^{2} + 12xy} =\]

\[= \frac{(x + 3y) \cdot (x + 3y)}{4x \cdot (x + 3y)} = \frac{x + 3y}{4x}\]

\[\frac{- 0,2 + 3 \cdot ( - 0,6)}{4 \cdot ( - 0,2)} =\]

\[= \frac{- 0,2 - 1,8}{- 0,8} = \frac{2}{0,8} = 2,5.\]