To solve the equation:

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

let's denote $a = \sqrt{2 - \sqrt{3}}$ and $b = \sqrt{2 + \sqrt{3}}$. Thus, the equation can be rewritten as:

$a^x + b^x = 4$

First, let's check if $x = 2$ is a solution.

Calculating $a^2$:

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

Similarly, calculating $b^2$:

$b^2 = (\sqrt{2 + \sqrt{3}})^2 = 2 + \sqrt{3}$

Now, sum these values:

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

So, $x = 2$ satisfies the equation.

Therefore, the solution to the equation is:

$x = 2$

Would you like more details or have any questions about this solution?

Here are 5 numbered questions you might ask next:

1. Can you show the steps to verify if $x = 2$ satisfies the original equation in more detail?
2. How can we solve similar exponential equations?
3. What are some common techniques for solving equations involving square roots?
4. How would this problem change if the constants inside the square roots were different?
5. Can you explain the properties of exponents used in this solution?

Tip: When dealing with equations involving square roots and exponents, always consider checking simple values for $x$ first, as they can often lead to straightforward solutions.