188° Вправа 188Алгебра 8 клас Тарасенкова 2025 · § 5 · стор. 57

188. Виконайте множення виразів.

Спершу розкладаємо чисельники і знаменники на:

1) ${\displaystyle (x-3)\cdot \frac{11}{x-3}=}$${\displaystyle \frac{\cancel{\left(x-3\right)}\cdot 11}{\cancel{x-3}}=}$${\displaystyle 11}$;

2) ${\displaystyle (3x+2)\cdot (-\frac{2}{2+3x})=}$${\displaystyle -\frac{\cancel{\left(3x+2\right)}\cdot 2}{\cancel{3x+2}}=}$${\displaystyle -2}$;

3) ${\displaystyle (2a-1)\cdot \frac{5}{1-2a}=}$${\displaystyle \frac{\cancel{\left(2a-1\right)}\cdot 5}{-\cancel{\left(2a-1\right)}}=}$${\displaystyle -5}$;

4) ${\displaystyle (2x-2)\cdot \frac{6}{x-1}=}$${\displaystyle \frac{2\cancel{\left(x-1\right)}\cdot 6}{\cancel{x-1}}=}$${\displaystyle 12}$;

5) ${\displaystyle (3a+1)\cdot \frac{5}{2+6a}=}$${\displaystyle \frac{\cancel{\left(3a+1\right)}\cdot 5}{2\cancel{\left(1+3a\right)}}=}$${\displaystyle 2,5}$;

6) ${\displaystyle (ax-a)\cdot \frac{2}{x-1}=}$${\displaystyle \frac{a\cancel{\left(x-1\right)}\cdot 2}{\cancel{x-1}}=}$${\displaystyle 2a}$;

7) ${\displaystyle (ax+2a)\cdot \frac{a}{x+2}=}$${\displaystyle \frac{a\cancel{\left(x+2\right)}\cdot a}{\cancel{x+2}}=}$${\displaystyle a^2}$;

8) ${\displaystyle (3b-ab)\cdot \frac{3}{2a-6}=}$${\displaystyle \frac{-b\cancel{\left(a-3\right)}\cdot 3}{2\cancel{\left(a-3\right)}}=}$${\displaystyle -\frac{3b}{2}=}$${\displaystyle -1,5b}$.