To find the value of \(x\) from the given equation:

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

We can cross-multiply to eliminate the fractions:

\(\sqrt{2} \cdot (x - \sqrt{2}) = 1 \cdot (x + \sqrt{2})\)

Expand both sides:

\(\sqrt{2}x - 2 = x + \sqrt{2}\)

Now, isolate the term with \(x\) on one side:

\(\sqrt{2}x - x = \sqrt{2} + 2\)

Factor out \(x\):

\(x (\sqrt{2} - 1) = \sqrt{2} + 2\)

Finally, solve for \(x\):

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

Rationalize the denominator:

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

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

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

So, the value of \(x\) is \(4 + 3\sqrt{2}\).