If G be the geometric mean of x and y, then f | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) As given $G = \sqrt {xy} $

$	herefore $$\frac{1}{{{G^2} - {x^2}}} + \frac{1}{{{G^2} - {y^2}}} = \frac{1}{{xy - {x^2}}} + \frac{1}{{xy - {y^2}}

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) As given $G = \sqrt {xy} $

$	herefore $$\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}}\left\{ { - \frac{1}{x} + \frac{1}{y}} ight\} = \frac{1}{{xy}} = \frac{1}{{{G^2}}}$.

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

Similar Questions

The sum of the series $5.05 + 1.212 + 0.29088 + ...\,\infty $ is

If $\frac{{x + y}}{2},\;y,\;\frac{{y + z}}{2}$ are in $H.P.$, then $x,\;y,\;z$ are in

The product of three geometric means between $4$ and $\frac{1}{4}$ will be

If $a, b, c$ and $d$ are in $G.P.$ show that:

$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$

If the first term of a $G.P.$ be $5$ and common ratio be $ - 5$, then which term is $3125$