Решебник по алгебре 8 класс Макарычев ФГОС Задание 27

 

\[\boxed{\text{27\ (27).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\frac{8^{16}}{16^{12}} = \frac{{(2^{3})}^{16}}{{(2^{4})}^{12}} = \frac{2^{48}}{2^{48}} = 1\]

\[\textbf{б)}\frac{81^{25}}{27^{33}} = \frac{\left( 3^{4} \right)^{25}}{\left( 3^{3} \right)^{33}} = \frac{3^{100}}{3^{99}} =\]

\[= 3^{1} = 3\]

\[\boxed{\text{27.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 4a^{2}b^{3}\ :\left( 2a^{4}b^{2} \right) = \frac{4a^{2}b^{3}}{2a^{4}b^{2}} =\]

\[= \frac{2a^{2}b^{2} \cdot 2b}{2a^{2}b^{2} \cdot a^{2}} = \frac{2b}{a^{2}}\]

\[\textbf{б)}\ 3xy^{2}\ :\left( 6x^{3}y^{3} \right) = \frac{3xy^{2}}{6x^{3}y^{3}} =\]

\[= \frac{3xy^{2}}{3xy^{2} \cdot 2x^{2}y} = \frac{1}{2x^{2}y}\]

\[\textbf{в)}\ 24\ {p^{4}q}^{4}\ :\left( 48p^{2}q^{2} \right) =\]

\[= \frac{24{p^{4}q}^{4}}{48p^{2}q^{2}} = \frac{24p^{2}q^{2} \cdot p^{2}q^{2}}{2 \cdot 24p^{2}q^{2}} =\]

\[= \frac{p^{2}q^{2}}{2}\]

\[\textbf{г)}\ 36\ m^{2}n\ \ :\left( 18\text{mn} \right) = \frac{36m^{2}n}{18mn} =\]

\[= \frac{2 \cdot 18mn \cdot m}{18mn} = 2m\]

\[\textbf{д)} - 32\ b^{5}\text{c\ }:\left( 12b^{4}c^{2} \right) =\]

\[= \frac{- 32\ b^{5}c}{12b^{4}c^{2}} = - \frac{8 \cdot 4b^{4}c \cdot b}{3 \cdot 4b^{4}c \cdot c} =\]

\[= - \frac{8b}{3c}\]

\[\textbf{е)} - 6ax\ \ :( - 18ax) = \frac{- 6ax}{- 18ax} =\]

\[= \frac{6ax}{3 \cdot 6ax} = \frac{1}{3}\]