Решебник по алгебре 8 класс Макарычев ФГОС Задание 73

 

\[\boxed{\text{73\ (73).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[= \frac{3x}{6} + \frac{2y}{6} = \frac{3x + 2y}{6}\ \]

\[= \frac{3c}{12} - \frac{d}{12} = \frac{3c - d}{12}\]

\[= \frac{a^{2} - b^{3}}{\text{ab}}\]

\[= \frac{9 - 4}{6x} = \frac{5}{6x}\]

\[= \frac{5x + 2x}{8y} = \frac{7x}{8y}\]

\[\textbf{е)}\ \frac{17y^{\backslash 3}}{24c} - \frac{25y^{\backslash 2}}{36c} =\]

\[= \frac{51y - 50y}{72c} = \frac{y}{72c}\]

\[= \frac{5 - 8}{25a} = - \frac{3}{25a}\]

\[\textbf{з)}\ \frac{3b^{\backslash b}}{4c} + \frac{c^{\backslash 2c}}{2b} =\]

\[\boxed{\text{73.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 8x^{4} - 16x^{3}y = 8x³ \cdot (x - 2y)\]

\[\textbf{б)}\ 15xy^{5} + 10y^{2} =\]

\[= 5y^{2} \cdot (3xy^{3} + 2)\]

\[\textbf{в)}\ 8a^{2} - 50y^{2} =\]

\[= 2 \cdot \left( 4a^{2} - 25y^{2} \right) =\]

\[= 2 \cdot (2a - 5y) \cdot (2a + 5y)\]

\[\textbf{г)}\ 18\ b^{2} - 98a^{2} =\]

\[= 2 \cdot \left( 9b^{2} - 49a^{2} \right) =\]

\[= 2 \cdot (3b - 7a) \cdot (3b + 7a)\]

\[\textbf{д)}\ x^{3} - 125 =\]

\[= (x - 5) \cdot (x^{2} + 5x + 25)\]

\[\textbf{е)}\ y^{3} + 8 =\]

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

\[\textbf{ж)}\ ab + 8a + 9b + 72 =\]

\[= a \cdot (b + 8) + 9 \cdot (b + 8) =\]

\[= (a + 9) \cdot (b + 8)\]

\[\textbf{з)}\ 6m - 12 - 2n + mn =\]

\[= m \cdot (6 + n) - 2 \cdot (6 + n) =\]

\[= (m - 2) \cdot (6 + n)\]