$$ \left(\sqrt{x}\right.+\sqrt2 $$) ($$ \sqrt{x}-\sqrt2 $$)

To simplify the expression 

\[
\left(\sqrt{x} + \sqrt{2}\right) \left(\sqrt{x} - \sqrt{2}\right),
\]

we can use the difference of squares formula. The difference of squares formula states that 

\[
(a + b)(a - b) = a^2 - b^2.
\]

Here, $a = \sqrt{x}$ and $b = \sqrt{2}$. Applying the formula:

\[
(\sqrt{x} + \sqrt{2})(\sqrt{x} - \sqrt{2}) = (\sqrt{x})^2 - (\sqrt{2})^2.
\]

Now, calculate each square:

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

Putting these values back into the expression gives:

\[
x - 2.
\]

### Final Answer:

\[
\left(\sqrt{x} + \sqrt{2}\right)\left(\sqrt{x} - \sqrt{2}\right) = x - 2.
\]

Would you like to go through more examples or do you have any questions about this?

Here are 5 related questions that might help you understand more:

1. How do you simplify expressions involving the sum and difference of cube roots?
2. What happens when you expand $(a + b)^2$ and $(a - b)^2$?
3. How would you simplify $\left(\sqrt{a} + \sqrt{b}\right)\left(\sqrt{a} - \sqrt{b}\right)$ for any positive $a$ and $b$?
4. What is the general approach to multiplying binomials containing square roots?
5. How can you use conjugates to rationalize the denominator of a fraction?

**Tip:** Using the difference of squares formula is a quick way to simplify expressions like $(a + b)(a - b)$, where the result is just the difference of the squares of the terms.

Learn how to simplify the expression (sqrt{x} + sqrt{2})(sqrt{x} - sqrt{2}) using the difference of squares formula. This problem covers algebraic concepts involving square roots and is suitable for Grades 7-9.

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