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МатематикаВопрос:
06.5. a) \(\frac{6}{x-y}-\frac{5}{x+y}\)\(\cdot\) \(\frac{x-y}{x+11y}\);
Ответ:
Преобразуем выражение:
- $$\frac{6}{x-y}-\frac{5}{x+y} = \frac{6(x+y)-5(x-y)}{(x-y)(x+y)} = \frac{6x+6y-5x+5y}{(x-y)(x+y)} = \frac{x+11y}{(x-y)(x+y)}$$
Умножим первую дробь на вторую:
$$\frac{x+11y}{(x-y)(x+y)} \cdot \frac{x-y}{x+11y} = \frac{1}{x+y}$$
Ответ: $$\frac{1}{x+y}$$
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