99° Вправа 99Алгебра 8 клас Тарасенкова 2025 · § 2 · стор. 31

99. Скоротіть дріб.

1) ${\displaystyle \frac{3x^3}{x^3-x^2}=}$${\displaystyle \frac{\cancel{x^2}\cdot 3x}{\cancel{x^2}(x-1)}=}$${\displaystyle \frac{3x}{x-1}}$;

2) ${\displaystyle \frac{4y^3-4y^4}{2y^5}=}$${\displaystyle \frac{\cancel{2y^3}\cdot 2(1-y)}{\cancel{2y^3}\cdot y^2}=}$${\displaystyle \frac{2\left(1-y\right)}{y^2}}$;

3) ${\displaystyle \frac{3a^2-6a}{12a^2}=}$${\displaystyle \frac{\cancel{3a}(a-2)}{\cancel{3a}\cdot 4a}=}$${\displaystyle \frac{a-2}{4a}}$;

4) ${\displaystyle \frac{5x^3-5x^2}{15x}=}$${\displaystyle \frac{\cancel{5x}\cdot x(x-1)}{\cancel{5x}\cdot 3}=}$${\displaystyle \frac{x\left(x-1\right)}{3}}$;

5) ${\displaystyle \frac{3y^3-3y}{9y^3}=}$${\displaystyle \frac{\cancel{3y}(y^2-1)}{\cancel{3y}\cdot 3y^2}=}$${\displaystyle \frac{y^2-1}{3y^2}}$;

6) ${\displaystyle \frac{3a^3+12a^2}{6a^2}=}$${\displaystyle \frac{\cancel{3a^2}(a+4)}{\cancel{3a^2}\cdot 2}=}$${\displaystyle \frac{a+4}{2}}$.