Question 121

The sum of $$\frac{1}{x + y}$$ and $$\frac{1}{x - y}$$ is

Solution

$$\ \frac{\ 1}{x+y}+\ \frac{\ 1}{x-y}$$=$$\ \frac{\ x-y+x+y}{x^2-y^2}$$=$$\ \frac{\ 2x}{x^2-y^2}$$                                                    


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