Решебник по алгебре 8 класс Мерзляк ФГОС Задание 87

\[\boxed{\text{87\ (87).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\frac{x}{y} = 4\]

\[1)\ \frac{y}{x} = \frac{1}{4}\]

\[2)\ \frac{2x - 3y}{y} = \frac{2x}{y} - \frac{3y}{y} =\]

\[= \frac{2x}{y} - 3 = 2 \cdot \frac{x}{y} - 3 = 2 \cdot 4 - 3 =\]

\[= 8 - 3 = 5\]

\[3)\ \frac{x^{2} + y^{2}}{\text{xy}} = \frac{x^{2}}{\text{xy}} + \frac{y^{2}}{\text{xy}} = \frac{x}{y} + \frac{y}{x} =\]

\[= 4 + \frac{1}{4} = 4\frac{1}{4} = 4,25\]

\[\boxed{\text{87.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{n + 6}{n} = \frac{n}{n} + \frac{6}{n} = 1 + \frac{6}{n} \Longrightarrow\]

\[\Longrightarrow целое\ \ при\ n = 1,\ 2,\ 3,\ 6.\]

\[2)\ \frac{3n^{2} - 4n - 14}{n} =\]

\[= \frac{3n^{2}}{n} - \frac{4n}{n} - \frac{14}{n} =\]

\[= 3n - 4 - \frac{14}{n} \Longrightarrow\]

\[целое,\ при\ n = 1,\ 2,\ 7,\ 14\]

\[3)\frac{4n + 7}{2n - 3} = \frac{4n - 6 + 13}{2n - 3} =\]

\[= \frac{4n - 6}{2n - 3} + \frac{13}{2n - 3} =\]

\[= \frac{2 \cdot (2n - 3)}{2n - 3} + \frac{13}{2n - 3} =\]

\[= 2 + \frac{13}{2n - 3} \Longrightarrow целое\ при\ n =\]

\[= 1,\ 2,\ 8.\]