The problem is to simplify the expression $\frac{\sqrt{2}+1}{\sqrt{2}-1}$.

1. Problem Description

The problem is to simplify the expression $\frac{\sqrt{2}+1}{\sqrt{2}-1}$.

2. Solution Steps

To simplify the expression, we need to rationalize the denominator. We multiply both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{2}+1$.

$\frac{\sqrt{2}+1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}$

Now we multiply the numerators and the denominators:

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

The numerator is $(\sqrt{2}+1)^2 = (\sqrt{2})^2 + 2(\sqrt{2})(1) + 1^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}$.

The denominator is $(\sqrt{2}-1)(\sqrt{2}+1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1$.

Therefore, the expression becomes:

$\frac{3 + 2\sqrt{2}}{1} = 3 + 2\sqrt{2}$

3. Final Answer

$3+2\sqrt{2}$