ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 51

 

\[\boxed{\text{51\ (51)\ .\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 5bc - 5c = 5c \cdot (b - 1)\]

\[\textbf{б)}\ 10n + 15n^{2} = 5n \cdot (2 + 3n)\]

\[\textbf{в)}\ 8ab + 12bc = 4b \cdot (2a + 3c)\]

\[\textbf{г)}\ 5y - 5x + y^{2} - xy =\]

\[= 5 \cdot (y - x) + y \cdot (y - x) =\]

\[= (y - x) \cdot (5 + y)\]

\[\textbf{д)}\ a^{2} - 9 = (a - 3) \cdot (a + 3)\]

\[\textbf{е)}\text{\ x}^{2} + 10x + 25 = ({x + 5)}^{2} =\]

\[= (x + 5)(x + 5)\]

\[\textbf{ж)}\ y^{2} - 2y + 1 = ({y - 1)}^{2} =\]

\[= (y - 1)(y - 1)\]

\[\textbf{з)}\text{\ a}^{3} + 64 =\]

\[= (a + 4) \cdot (a^{2} - 4a + 16)\]

\[\textbf{и)}\ b^{3} - 1 =\]

\[= (b - 1) \cdot (b^{2} + b + 1)\]

\(\boxed{\text{51.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\)

Пояснение.

Решение.

\[\textbf{а)}\frac{x}{a - b} = \frac{x \cdot (a - b)}{(a - b) \cdot (a - b)} =\]

\[= \frac{xa - xb}{(a - {b)}^{2}}\]

\[\textbf{б)}\ \frac{y}{x - a} = \frac{y \cdot (x + a)}{(x - a) \cdot (x + a)} =\]

\[= \frac{yx + ya}{x^{2} - a^{2}}\]

\[\textbf{в)}\ \frac{a}{a - 10} = \frac{a \cdot ( - 1)}{(a - 10) \cdot ( - 1)} =\]

\[= \frac{- a}{10 - a} = - \frac{a}{10 - a}\]

\[\textbf{г)}\ \frac{p}{p - 2} =\]

\[= \frac{p \cdot \left( - (p + 2) \right)}{(p - 2) \cdot \left( - (p + 2) \right)} =\]

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

\[\textbf{д)}\ \frac{\text{mn}}{n - m} = - \frac{\text{mn}(m + n)}{(m - n)(m + n)} =\]

\[= - \frac{mn(m + n)}{m^{2} - n^{2}}\]