If G is the geometric mean of x and y,then fr | vedclass

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(B) Given that $G$ is the geometric mean of $x$ and $y$,we have $G = \sqrt{xy}$,which implies $G^2 = xy$.Substituting $G^2 = xy$ into the expression:$

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$\frac{1}{G^2}$

Vedclass · Answer

(B) Given that $G$ is the geometric mean of $x$ and $y$,we have $G = \sqrt{xy}$,which implies $G^2 = xy$.Substituting $G^2 = xy$ into the expression:$\frac{1}{G^2 - x^2} + \frac{1}{G^2 - y^2} = \frac{1}{xy - x^2} + \frac{1}{xy - y^2}$$= \frac{1}{x(y - x)} + \frac{1}{y(x - y)}$$= \frac{1}{x(y - x)} - \frac{1}{y(y - x)}$$= \frac{1}{y - x} \left( \frac{1}{x} - \frac{1}{y} \right)$$= \frac{1}{y - x} \left( \frac{y - x}{xy} \right)$$= \frac{1}{xy} = \frac{1}{G^2}$.

If $G$ is the geometric mean of $x$ and $y$,then $\frac{1}{G^2 - x^2} + \frac{1}{G^2 - y^2} = $

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