If `x gt2` then what is the value of `sqrt(x+2sqrt(x-1))+sqrt(x-2sqrtx-1)`?

Options

A. 2
B. `2sqrt(x-1)`
C. `-2`
D. `-2sqrt(x-1)`

Answer

To solve the expression \( \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} \) given that \( x > 2 \), we can simplify each square root term step by step.

### Step 1: Simplify the first term
We start with the first term:
\[ \sqrt{x + 2\sqrt{x - 1}} \]
Notice that \( x \) can be rewritten as \( \sqrt{x}^2 \). Therefore, we can express the term as:
\[ \sqrt{\left(\sqrt{x}\right)^2 + 2\sqrt{x - 1}} \]
This resembles the expansion of a binomial square, \( (a + b)^2 = a^2 + 2ab + b^2 \). We can rewrite it as:
\[ \sqrt{(\sqrt{x} + \sqrt{x - 1})^2} \]
Thus, we have:
\[ \sqrt{x + 2\sqrt{x - 1}} = \sqrt{x} + \sqrt{x - 1} \]

### Step 2: Simplify the second term
Now, we simplify the second term:
\[ \sqrt{x - 2\sqrt{x - 1}} \]
Similarly, we can express this as:
\[ \sqrt{\left(\sqrt{x}\right)^2 - 2\sqrt{x - 1}} \]
This also resembles the binomial square, but in this case, it can be rewritten as:
\[ \sqrt{(\sqrt{x} - \sqrt{x - 1})^2} \]
Thus, we have:
\[ \sqrt{x - 2\sqrt{x - 1}} = \sqrt{x} - \sqrt{x - 1} \]

### Step 3: Combine the results
Now we can combine both simplified terms:
\[ \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} = (\sqrt{x} + \sqrt{x - 1}) + (\sqrt{x} - \sqrt{x - 1}) \]
This simplifies to:
\[ 2\sqrt{x} \]

### Conclusion
Therefore, the value of the expression \( \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} \) when \( x > 2 \) is:
\[ \boxed{2\sqrt{x}} \]