Let's solve for the given expression:
Given that \( x = 3 - 2\sqrt{2} \), we need to find \( \frac{\sqrt{x+1}}{\sqrt{x}} \).
Using the provided hint:
\[ (\sqrt{x+1})^2 = x + \frac{1}{x} + 2 \]
Substitute \( x = 3 - 2\sqrt{2} \) into the equation:
\[ (\sqrt{3-2\sqrt{2}+1})^2 = 3 - 2\sqrt{2} + \frac{1}{3-2\sqrt{2}} + 2 \]
Simplify the expression under the square root:
\[ (\sqrt{4-2\sqrt{2}})^2 = 3 - 2\sqrt{2} + \frac{1}{3-2\sqrt{2}} + 2 \]
\[ 4 - 2\sqrt{2} = 3 - 2\sqrt{2} + \frac{1}{3-2\sqrt{2}} + 2 \]
Now, subtract \( 3 - 2\sqrt{2} + 2 \) from both sides:
\[ 4 - 2\sqrt{2} - 3 + 2\sqrt{2} - 2 = \frac{1}{3-2\sqrt{2}} \]
\[ -1 = \frac{1}{3-2\sqrt{2}} \]
Now, multiply both sides by \( 3-2\sqrt{2} \) to isolate the denominator:
\[ -(3-2\sqrt{2}) = 1 \]
\[ 2\sqrt{2} - 3 = 1 \]
Add 3 to both sides:
\[ 2\sqrt{2} = 4 \]
Divide by 2:
\[ \sqrt{2} = 2 \]
Now, substitute this value back into the original expression \( \frac{\sqrt{x+1}}{\sqrt{x}} \):
\[ \frac{\sqrt{3 - 2\sqrt{2} + 1}}{\sqrt{3 - 2\sqrt{2}}} = \frac{\sqrt{2}}{\sqrt{2}} = 1 \]
Therefore, the value of \( \frac{\sqrt{x+1}}{\sqrt{x}} \) is 1.
