Решебник по алгебре 8 класс Макарычев ФГОС Задание 214

 

\[\boxed{\text{214\ (214).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{б)}\ \frac{\left( a^{2} - 9 \right)^{2}}{(3 - a)^{3}} =\]

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

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

\[\textbf{в)}\ \frac{8y^{3} - 1}{y - 4y^{3}} =\]

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

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

\[= - \frac{4y^{2} + 2y + 1}{y(1 + 2y)}\]

\[\textbf{г)}\ \frac{5a^{2} - 3ab}{a^{2} - 0,36b^{2}} = \frac{a(5a - 3b)}{a^{2} - 0,36b^{2}} =\]

\[= \frac{5a}{a + 0,6b}\ \]

\[\boxed{\text{214.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{3x - 28}{25}\]

\[x \in R.\]

\[\textbf{б)}\ \frac{37}{2y + 7}\]

\[2y + 7 \neq 0\]

\[2y \neq - 7\]

\[y \neq - 3,5.\]

\[\textbf{в)}\ \frac{9}{x^{2} - 7x}\]

\[x^{2} - 7x \neq 0\]

\[x(x - 7) \neq 0\]

\[x \neq 0\]

\[x \neq 7.\]

\[\textbf{г)}\ \frac{2y + 5}{y^{2} + 8}\]

\[y^{2} + 8 \neq 0\]

\[y^{2} \neq - 8\]

\[y \in R.\]

\[\textbf{д)}\ \frac{12}{|x| - 3}\]

\[|x| - 3 \neq 0\]

\[|x| \neq 3\]

\[x \neq \pm 3.\]

\[\textbf{е)}\ \frac{45}{|y| + 2}\]

\[|y| + 2 \neq 0\]

\[y \in R.\]