9 Вправа 9 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 317

9. Виконайте дії:

1) ${\displaystyle \frac{\frac{1}{a+b}-\frac{1}{a-b}}{\frac{1}{a+b}+\frac{1}{a-b}}\cdot \frac{a}{b}=}$${\displaystyle \frac{\frac{a-b-(a+b)}{(a+b)(a-b)}}{\frac{a-b+a+b}{(a+b)(a-b)}}\cdot \frac{a}{b}=}$${\displaystyle \frac{\cancel{a}-b\cancel{-a}-b}{\cancel{\left(a+b\right)\left(a-b\right)}}\cdot \frac{\cancel{\left(a+b\right)\left(a-b\right)}}{\cancel{a}-b+\cancel{a}+b}\cdot \frac{a}{b}=}$${\displaystyle \frac{-\stackrel{1}{\cancel{2}}\cancel{b}}{\underset{1}{\cancel{2}}\cancel{a}}\cdot \frac{\cancel{a}}{\cancel{b}}=}$${\displaystyle -1}$;

2) ${\displaystyle \frac{\frac{1}{a}+\frac{1}{b+c}}{\frac{1}{a}-\frac{1}{b+c}}\cdot (1+\frac{b^2+c^2-a^2}{2bc})=}$${\displaystyle \frac{\frac{b+c+a}{\cancel{a\left(b+c\right)}}}{\frac{b+c-a}{\cancel{a\left(b+c\right)}}}\cdot \frac{2bc+b^2+c^2-a^2}{2bc}=}$${\displaystyle \frac{a+b+c}{b+c-a}\cdot \frac{{(b+c)}^2-a^2}{2bc}=}$${\displaystyle \frac{a+b+c}{\cancel{b+c-a}}\cdot \frac{\cancel{\left(b+c-a\right)}(b+c+a)}{2bc}=}$${\displaystyle \frac{{\left(a+b+c\right)}^2}{2bc}}$.