If frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k left | vedclass

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(D) Given the equation: $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} \right)$ Simplify the right side

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$\frac{1}{a^2 - b^2}$

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(D) Given the equation: $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} \right)$ Simplify the right side by taking a common denominator: $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2(x^2 + b^2) - b^2(x^2 + a^2)}{(x^2 + a^2)(x^2 + b^2)} \right)$ Equating the numerators: $x^2 = k [a^2 x^2 + a^2 b^2 - b^2 x^2 - a^2 b^2]$ $x^2 = k [a^2 x^2 - b^2 x^2]$ $x^2 = k x^2 (a^2 - b^2)$ Dividing both sides by $x^2$: $1 = k(a^2 - b^2)$ Therefore,$k = \frac{1}{a^2 - b^2}$.

If $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} \right)$,then $k =$

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