139° Вправа 139Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 45

139. Виконайте додавання дробів.

1) ${\displaystyle \frac{2}{5x}+\frac{7}{5x}=}$${\displaystyle \frac{2+7}{5x}=}$${\displaystyle \frac{9}{5x}}$;

2) ${\displaystyle \frac{8}{3a}+\frac{2}{3a}=}$${\displaystyle \frac{8+2}{3a}=}$${\displaystyle \frac{10}{3a}}$;

3) ${\displaystyle \frac{1}{6c}+\frac{7}{6c}=}$${\displaystyle \frac{1+7}{6c}=}$${\displaystyle \frac{8}{6c}=}$${\displaystyle \frac{\stackrel{4}{\cancel{8}}}{\underset{3}{\cancel{6}}c}=}$${\displaystyle \frac{4}{3c}}$;

4) ${\displaystyle \frac{2x}{x-2}+\frac{3x}{x-2}=}$${\displaystyle \frac{2x+3x}{x-2}=}$${\displaystyle \frac{5x}{x-2}}$;

5) ${\displaystyle \frac{3y}{y+1}+\frac{2y}{y+1}=}$${\displaystyle \frac{3y+2y}{y+1}=}$${\displaystyle \frac{5y}{y+1}}$;

6) ${\displaystyle \frac{4c}{c-1}+\frac{c}{c-1}=}$${\displaystyle \frac{4c+c}{c-1}=}$${\displaystyle \frac{5c}{c-1}}$;

7) ${\displaystyle \frac{3a}{x+2}+\frac{4a}{x+2}=}$${\displaystyle \frac{3a+4a}{x+2}=}$${\displaystyle \frac{7a}{x+2}}$;

8) ${\displaystyle \frac{4c}{y-1}+\frac{7c}{y-1}=}$${\displaystyle \frac{4c+7c}{y-1}=}$${\displaystyle \frac{11c}{y-1}}$;

9) ${\displaystyle \frac{3x}{c^2+1}+\frac{8x}{c^2+1}=}$${\displaystyle \frac{3x+8x}{c^2+1}=}$${\displaystyle \frac{11x}{c^2+1}}$.