x= −3 2√2 ,then √x+1/√x is equal to

The correct option is 2√2.

Step-by-step Derivation:

First, we note that the given expression for $x$ contains a typographical error in the question statement ($x=-32\sqrt{2}$). The standard form of this mathematical problem is:
$x=3-2\sqrt{2}$

We need to find the value of the expression:
$\sqrt{x}+\frac{1}{\sqrt{x}}$

To do this, let us first simplify $\sqrt{x}$ by expressing $x$ as a perfect square. We can rewrite the number 3 as $2+1$:
$x=2+1-2\sqrt{2}$

This can be written in the form of the algebraic identity $a^2+b^2-2ab={(a-b)}^2$:
$x={\left(\sqrt{2}\right)}^2+1^2-2\left(\sqrt{2}\right)\left(1\right)$

Applying the identity where $a=\sqrt{2}$ and $b=1$:
$x={(\sqrt{2}-1)}^2$

Taking the square root on both sides (since $\sqrt{2}>1$, the positive root is):
$\sqrt{x}=\sqrt{2}-1$

Now, we find the reciprocal of $\sqrt{x}$:
$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{2}-1}$

To simplify this fraction, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate, $\sqrt{2}+1$:
$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

Using the difference of squares formula, $(u-v)(u+v)=u^2-v^2$, in the denominator:
$\frac{1}{\sqrt{x}}=\frac{\sqrt{2}+1}{{\left(\sqrt{2}\right)}^2-1^2}=\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1$

Finally, we add the two expressions together:
$\sqrt{x}+\frac{1}{\sqrt{x}}=(\sqrt{2}-1)+(\sqrt{2}+1)$

Simplifying the sum:
$\sqrt{x}+\frac{1}{\sqrt{x}}=2\sqrt{2}$