Answer
$\dfrac{5+\sqrt{2}}{\sqrt{2x}}=\dfrac{23}{5\sqrt{2x}-2\sqrt{x}}$
Work Step by Step
$\dfrac{5+\sqrt{2}}{\sqrt{2x}}$
Multiply the numerator and the denominator by the conjugate of the numerator and simplify if possible:
$\dfrac{5+\sqrt{2}}{\sqrt{2x}}=\dfrac{5+\sqrt{2}}{\sqrt{2x}}\cdot\dfrac{5-\sqrt{2}}{5-\sqrt{2}}=\dfrac{5^{2}-(\sqrt{2})^{2}}{\sqrt{2x}(5-\sqrt{2})}=...$
$...=\dfrac{25-2}{5\sqrt{2x}-\sqrt{4x}}=\dfrac{23}{5\sqrt{2x}-2\sqrt{x}}$
