Решебник по алгебре 8 класс Макарычев ФГОС Задание 26

\[\boxed{\text{26\ (26).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{4a^{2}}{6ac} = \frac{2 \cdot 2a \cdot a}{3 \cdot 2a \cdot c} = \frac{2a}{3c}\]

\[\textbf{б)}\frac{7x^{2}y}{21xy^{2}} = \frac{7xy \cdot x}{3 \cdot 7xy \cdot y} = \frac{x}{3y}\]

\[\textbf{в)}\ \frac{56m^{2}n^{5}}{35mn^{5}} = \frac{8 \cdot 7mn^{5} \cdot m}{5 \cdot 7mn^{5}} = \frac{8m}{5}\]

\[\textbf{г)}\frac{25p^{4}q}{100p^{5}q} = \frac{25p^{4}q}{4 \cdot 25p^{4}q \cdot p} = \frac{1}{4p}\]

\[\boxed{\text{26.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\frac{10\text{xz}}{15\text{yz}} = \frac{5z \cdot 2x}{5z \cdot 3y} = \frac{2x}{3y}\text{\ \ }\]

\[\textbf{б)}\frac{6ab^{2}}{9bc^{2}} = \frac{3b \cdot 2ab}{3b \cdot 3c²} = \frac{2ab}{3c^{2}}\]

\[\textbf{в)}\frac{2ay^{3}}{- 4a^{2}b} = - \frac{2a \cdot y^{3}}{2a \cdot 2ab} = - \frac{y^{3}}{2ab}\]

\[\textbf{г)}\frac{- 6p^{2}q}{- 2q^{3}} = \frac{2q \cdot 3p^{2}}{2q \cdot q²} = \frac{3p^{2}}{q^{2}}\]

\[\textbf{д)}\frac{24a^{2}c^{2}}{36ac} = \frac{12ac \cdot 2ac}{12ac \cdot 3} = \frac{2ac}{3}\]

\[\textbf{е)}\frac{63x^{2}y^{3}}{42x^{6}y^{4}} = \frac{21x^{2}y^{3} \cdot 3}{21x^{2}y^{3} \cdot 2x^{4}y} =\]

\[= \frac{3}{2x^{4}y}\]