215° Вправа 215Алгебра 8 клас Тарасенкова 2025 · § 6 · стор. 64

215. Знайдіть частку дробів.

1) ${\displaystyle \frac{x+1}{3}:\frac{x+1}{15}=}$${\displaystyle \frac{x+1}{3}\cdot \frac{15}{x+1}=}$${\displaystyle \frac{\cancel{\left(x+1\right)}\cdot 5\cdot \stackrel{1}{\cancel{3}}}{\underset{1}{\cancel{3}}\cancel{\left(x+1\right)}}=}$${\displaystyle 5}$;

2) ${\displaystyle \frac{5y-5}{7}:\frac{y-1}{28}=}$${\displaystyle \frac{5(y-1)}{7}\cdot \frac{28}{y-1}=}$${\displaystyle \frac{5\cancel{\left(y-1\right)}\cdot 4\cdot \stackrel{1}{\cancel{7}}}{\underset{1}{\cancel{7}}\cancel{\left(y-1\right)}}=}$${\displaystyle 20}$;

3) ${\displaystyle \frac{2a+1}{8}:\frac{4a+2}{16}=}$${\displaystyle \frac{2a+1}{8}\cdot \frac{16}{2\left(2a+1\right)}=}$${\displaystyle \frac{\cancel{\left(2a+1\right)}\cdot \stackrel{1}{\cancel{16}}}{\underset{1}{\cancel{16}}\cancel{\left(2a+1\right)}}=}$${\displaystyle 1}$;

4) ${\displaystyle \frac{1-b}{5}:\frac{{(b-1)}^2}{25}=}$${\displaystyle \frac{1-b}{5}\cdot \frac{25}{{\left(1-b\right)}^2}=}$${\displaystyle \frac{\cancel{\left(1-b\right)}\cdot 5\cdot \stackrel{1}{\cancel{5}}}{\underset{1}{\cancel{5}}\cancel{\left(1-b\right)}(1-b)}=}$${\displaystyle \frac{5}{1-b}}$;

5) ${\displaystyle \frac{3-6y}{14}:\frac{2y-1}{42}=}$${\displaystyle \frac{-3(2y-1)}{14}\cdot \frac{42}{2y-1}=}$${\displaystyle \frac{-3\cancel{\left(2y-1\right)}\cdot 3\cdot \stackrel{1}{\cancel{14}}}{\underset{1}{\cancel{14}}\cancel{\left(2y-1\right)}}=}$${\displaystyle -9}$;

6) ${\displaystyle \frac{1-3x}{18}:\frac{9x-3}{42}=}$${\displaystyle \frac{-(3x-1)}{18}\cdot \frac{42}{3\left(3x-1\right)}=}$${\displaystyle -\frac{\cancel{\left(3x-1\right)}\cdot \stackrel{7}{\cancel{42}}}{\underset{3}{\cancel{18}}\cdot 3\cancel{\left(3x-1\right)}}=}$${\displaystyle -\frac{7}{9}}$.

(У пункті 4) ${\displaystyle {(b-1)}^2=}$${\displaystyle {(1-b)}^2}$.)