ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 74

 

\[\boxed{\text{74\ (74).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{5y - 3^{\backslash 2}}{6y} + \frac{y + 2^{\backslash 3}}{4y} =\]

\[= \frac{2 \cdot (5y - 3) + 3 \cdot (y + 2)}{12y}\]

\[= \frac{10y - 6 + 3y + 6}{12y} = \ \frac{13y}{12y} =\]

\[= \frac{13}{12} = 1\frac{1}{12}\]

\[\textbf{б)}\ \frac{3x + 5^{\backslash 3}}{35x} + \frac{x - 3^{\backslash 5}}{21x} =\]

\[= \frac{9x + 15 + 5x - 15}{105x} =\]

\[= \frac{14x}{105x} = \frac{2}{15}\]

\[\textbf{в)}\ \frac{b + 2^{\backslash 3c}}{15b} - \frac{3c - 5^{\backslash b}}{45c} =\]

\[= \frac{3bc + 6c - 3bc + 5b}{45cb} =\]

\[= \frac{6c + 5b}{45bc}\]

\[\textbf{г)}\ \frac{8b + y^{\backslash 3y}}{40b} - \frac{6y + b^{\backslash 4b}}{30y} =\]

\[= \frac{24by + 3y^{2} - 24by - 4b^{2}}{120by} =\]

\[= \frac{3y^{2} - 4b^{2}}{120by}\]

\[\boxed{\text{74.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{3a}{2a + 25}\ \]

\[2a + 25 = 0\]

\[2a = - 25\]

\[a = - 12,5\]

\[Ответ:\ \ при\ a \neq - 12,5\]

\[\textbf{б)}\ \frac{2y}{9 + y^{2}}\]

\[9 + y^{2} = 0\]

\[y^{2} = - 9\]

\[y^{2}\ не\ может\ быть\ \]

\[отрицательным\ числом.\]

\[Ответ:при\ любых\ y.\]

\[\textbf{в)}\ \frac{5x}{3x \cdot (x + 12)}\]

\[3x \cdot (x + 12) = 0\]

\[3x = 0;\ \ \ \ x + 12 = 0\]

\[x = 0\ \ \ \ \ \ \ \ x = - 12\]

\[Ответ:при\ x \neq 0\ и\ x \neq - 12.\]

\[\textbf{г)}\ \frac{7a}{(a + 1) \cdot (a - 4)}\]

\[(a + 1) \cdot (a - 4) = 0\]

\[a + 1 = 0;\ \ \ \ \ a - 4 = 0\]

\[a = - 1\ \ \ \ \ \ \ \ \ \ \ a = 4.\]

\[Ответ:при\ a \neq - 1\ и\ a \neq 4.\]