268 Вправа 268Алгебра 8 клас Бевз 2025 · § 6 · стор. 58

268. Знайди частку.

а) ${\displaystyle \frac{1}{x+y}:\frac{1}{{(x+y)}^2}=}$${\displaystyle \frac{1}{x+y}\cdot {(x+y)}^2=}$${\displaystyle \frac{{(x+y)}^2}{x+y}}$=

=${\displaystyle \frac{\cancel{\left(x+y\right)}(x+y)}{\cancel{\left(x+y\right)}}=}$${\displaystyle x+y}$;

б) ${\displaystyle \frac{{(a+b)}^2}{3c}:\frac{a+b}{6c}=}$${\displaystyle \frac{{(a+b)}^2}{3c}\cdot \frac{6c}{a+b}=}$${\displaystyle \frac{{(a+b)}^2\cdot 6c}{3c\cdot (a+b)}}$=

=${\displaystyle \frac{2\cdot \cancel{3}\cancel{c}\cancel{\left(a+b\right)}(a+b)}{\cancel{3}\cancel{c}\cancel{\left(a+b\right)}}=}$${\displaystyle 2\left(a+b\right)}$;

в) ${\displaystyle \frac{x}{x+y}:\frac{x+y}{x}=}$${\displaystyle \frac{x}{x+y}\cdot \frac{x}{x+y}}$=

=${\displaystyle \frac{x\cdot x}{(x+y)\cdot (x+y)}=}$${\displaystyle \frac{x^2}{{\left(x+y\right)}^2}}$.