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\(\large\begin{array}{I} \mathtt{ \dfrac{ \sqrt{9x^2}}{ \sqrt{18y^2}}= \dfrac{3x}{ 3\sqrt{2}y}= \dfrac{x}{ \sqrt{2}y}}= \dfrac{ \sqrt{2}x}{2y} \end{array}\)

Answer is A
A3x /3ysqrtof2. Multiply numerator and denominator by sqrt of 2
Then you have 3xsqrt of 2 /. 6y reduce by 3 both numerator and denominator
You have x sqrt of 2 / 2y

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