Answer
Although $(\frac{2}{\sqrt x})^{2}\ne(\frac{2}{\sqrt x})\times(\frac{\sqrt x}{\sqrt x})$, if we simplify both of these expressions we will get identical expressions.
Work Step by Step
To determine whether the rationalization is the same as squaring this fraction, let's consider what happens in both of these cases:
Before making any changes to the base expression, we have to consider, that $x\gt0$, because square root of a number can't be negative, as well as denominator can't be zero. Now let's first square this fraction
$$(\frac{2}{\sqrt x})^{2}=\frac{4}{x} (1)$$
Now let's rationalize the expression
$$(\frac{2}{\sqrt x})\times(\frac{\sqrt x}{\sqrt x})=\frac{2\sqrt x}{x}(2)$$
By one look these expressions look different, but if we square the second expression we will get:
$$(\frac{2\sqrt x}{x})^2=\frac{4x}{x^2}=\frac{4}{x}(3)$$
So we get that the first expression is the same as the second expression (But note, we have to keep in mind that although we get $x$ without radical the base expression still forbids us to assign negative numbers to $x$)
