`2+sqrt2+1/(2+sqrt2)-1/(2-sqrt2)` is equal to

Options

A. 2
B. `2-sqrt2`
C. `4+sqrt2`
D. `2sqrt2`

Answer

To solve the expression \( 2 + \sqrt{2} + \frac{1}{2 + \sqrt{2}} - \frac{1}{2 - \sqrt{2}} \), we will follow these steps:

### Step 1: Rationalize the fractions
We will start by rationalizing the two fractions \( \frac{1}{2 + \sqrt{2}} \) and \( \frac{1}{2 - \sqrt{2}} \).

\[ \frac{1}{2 + \sqrt{2}} \cdot \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{2 - \sqrt{2}}{(2 + \sqrt{2})(2 - \sqrt{2})} \]

The denominator simplifies as follows:
\[ (2 + \sqrt{2})(2 - \sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2 \]

Thus, we have:
\[ \frac{1}{2 + \sqrt{2}} = \frac{2 - \sqrt{2}}{2} \]

Now for the second fraction:
\[ \frac{1}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{2 + \sqrt{2}}{(2 - \sqrt{2})(2 + \sqrt{2})} \]

The denominator simplifies similarly:
\[ (2 - \sqrt{2})(2 + \sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2 \]

Thus, we have:
\[ \frac{1}{2 - \sqrt{2}} = \frac{2 + \sqrt{2}}{2} \]

### Step 2: Substitute back into the expression
Now we can substitute these back into the original expression:
\[ 2 + \sqrt{2} + \frac{2 - \sqrt{2}}{2} - \frac{2 + \sqrt{2}}{2} \]

### Step 3: Combine the fractions
Combining the fractions gives:
\[ 2 + \sqrt{2} + \left(\frac{2 - \sqrt{2} - (2 + \sqrt{2})}{2}\right) \]
This simplifies to:
\[ 2 + \sqrt{2} + \frac{2 - \sqrt{2} - 2 - \sqrt{2}}{2} = 2 + \sqrt{2} + \frac{-2\sqrt{2}}{2} \]
\[ = 2 + \sqrt{2} - \sqrt{2} = 2 \]

### Final Answer
Thus, the expression \( 2 + \sqrt{2} + \frac{1}{2 + \sqrt{2}} - \frac{1}{2 - \sqrt{2}} \) simplifies to:
\[ \boxed{2} \]