Question

$\sqrt{ \frac{x}{y} } + \sqrt{ \frac{y}{x} } = \frac{10}{3} \\ find \: xy$
Answer plz. Answer is given 9 can anyone help me?

Answer 1

Iska answer 9 tb aayega agr x-y=8 esi kujh condition given ho saath mei

Answer 2

Answer:

$= > \sqrt{ \frac{x}{y} } + \sqrt{ \frac{y}{x} } = \frac{10}{3}$

$= > \frac{ \sqrt{x} \sqrt{x} + \sqrt{y} \sqrt{y} }{ \sqrt{xy} } = \frac{10}{3}$

$= > \frac{x + y}{ \sqrt{xy} } = \frac{10}{3}$

$= > \frac{3}{10} (x + y) = \sqrt{xy}$

$= > {(\frac{3}{10} (x + y)) }^{2} =xy$

$= > \frac{9}{100}{(x + y) }^{2} = xy$

$= > 9( {x}^{2} + 2xy + {y}^{2} ) = 100xy$

$= > 9 {x}^{2} + 18xy + 9 {y}^{2} = 100xy$

$= > 9 {x}^{2} + 9 {y}^{2} = 100xy - 18xy$

$= > 9 {x}^{2} + 9 {y}^{2} = 82xy$

$= > \frac{9}{82} ( {x}^{2} {y}^{2}) = xy$

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