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Among the following, a correct statement is $(w h e r e x > 0, y > 0)$

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$\sqrt{x . y} = \sqrt{x} . \sqrt{y}$

$\sqrt{x^{2} - y^{2}} = x - y$

$\sqrt{x^{2} + y^{2}} = x + y$

$\sqrt{x + y} = \sqrt{x} + \sqrt{y}$

answer is C.

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Detailed Solution

$\sqrt{x + y} \neq \sqrt{x} + \sqrt{y}$

$\sqrt{x^{2} + y^{2}} \neq x + y$

$\sqrt{x . y} = \sqrt{x} . \sqrt{y}$

$\sqrt{x^{2} - y^{2}} \neq x - y$

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