Question

For all x > 0 and y > 0, the radical expression:

Answer 1

$\frac{ \sqrt{x} }{3 \sqrt{x} - \sqrt{y} } = \frac{3x+ \sqrt{xy} }{9x-y} =\frac{3x+ \sqrt{xy} }{(3 \sqrt{x} - \sqrt{y})(3 \sqrt{x} + \sqrt{y})} \\ \sqrt{x} =\frac{ \sqrt{x} (3 \sqrt{x} + \sqrt{y}) }{(3 \sqrt{x} + \sqrt{y})} \\ \sqrt{x} = \sqrt{x} \\ 1=1$