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If $$\frac{p}{q}=\frac{x+3}{x-3}$$, then what is the value of $$\frac{p^2+q^2}{p^2-q^2}$$ ?
$$\frac{x^2+9}{3x}$$
$$\frac{x^2+18}{6x}$$
$$\frac{x^2+18}{3x}$$
$$\frac{x^2+9}{6x}$$
Given : $$\frac{p}{q}=\frac{x+3}{x-3}$$
Let $$p=(x+3)$$ and $$q=(x-3)$$
To find : $$\frac{p^2+q^2}{p^2-q^2}$$
= $$\frac{(x+3)^2+(x-3)^2}{(x+3)^2-(x-3)^2}$$
= $$\frac{(x^2+6x+9)+(x^2-6x+9)}{(x^2+6x+9)-(x^2-6x+9)}$$
= $$\frac{2x^2+18}{12x}$$
= $$\frac{x^2+9}{6x}$$
=> Ans - (D)
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