146° Вправа 146Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 47

146. Спростіть вираз.

Знаменник залишаємо тим самим, а чисельники додаємо (віднімаємо): ${\displaystyle \frac{A}{C}\pm \frac{B}{C}=}$${\displaystyle \frac{A\pm B}{C}}$. Отриманий дріб скорочуємо.

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

2) ${\displaystyle \frac{10}{5-x}-\frac{2x}{5-x}=}$${\displaystyle \frac{10-2x}{5-x}=}$${\displaystyle \frac{2\cancel{\left(5-x\right)}}{\cancel{5-x}}=}$${\displaystyle 2}$;

3) ${\displaystyle \frac{11}{6+c}+\frac{1+2c}{6+c}=}$${\displaystyle \frac{11+1+2c}{6+c}=}$${\displaystyle \frac{12+2c}{6+c}=}$${\displaystyle \frac{2\cancel{\left(6+c\right)}}{\cancel{6+c}}=}$${\displaystyle 2}$;

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

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

6) ${\displaystyle \frac{4c}{c-1}+\frac{1-5c}{c-1}=}$${\displaystyle \frac{4c+1-5c}{c-1}=}$${\displaystyle \frac{1-c}{c-1}=}$${\displaystyle \frac{-\cancel{\left(c-1\right)}}{\cancel{c-1}}=}$${\displaystyle -1}$.