Given:
$$x=7-4\sqrt{3}$$

We can express $7-4\sqrt{3}$ as a perfect square:

$$x=4+3-2(2)(\sqrt{3})=(2{)}^2+(\sqrt{3}{)}^2-2(2)(\sqrt{3})$$

$$x=(2-\sqrt{3}{)}^2$$

Taking the square root on both sides:

$$\sqrt{x}=2-\sqrt{3}$$

Now, find the reciprocal $\frac{1}{\sqrt{x}}$:

$$\frac{1}{\sqrt{x}}=\frac{1}{2-\sqrt{3}}$$

Rationalizing the denominator:

$$\frac{1}{\sqrt{x}}=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{2^2-(\sqrt{3}{)}^2}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$$

Adding the two values:

$$\sqrt{x}+\frac{1}{\sqrt{x}}=(2-\sqrt{3})+(2+\sqrt{3})=4$$