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МатематикаВопрос:
г) \(\frac{x^2-9}{2x^2+1}\)\(\cdot\) \(\frac{6x+1}{x-3}+\frac{6x-1}{x+3}\).
Ответ:
Преобразуем выражение:
- $$\frac{x^2-9}{2x^2+1} = \frac{(x-3)(x+3)}{2x^2+1}$$
- $$\frac{6x+1}{x-3}+\frac{6x-1}{x+3} = \frac{(6x+1)(x+3)+(6x-1)(x-3)}{(x-3)(x+3)} = \frac{6x^2+18x+x+3+6x^2-18x-x+3}{(x-3)(x+3)} = \frac{12x^2+6}{(x-3)(x+3)} = \frac{6(2x^2+1)}{(x-3)(x+3)}$$
Умножим первую дробь на вторую:
$$\frac{(x-3)(x+3)}{2x^2+1} \cdot \frac{6(2x^2+1)}{(x-3)(x+3)} = 6$$
Ответ: 6
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