Answer :
Answer:MARK AS BRAINLIST
Step-by-step explanation:
To find \( x + \frac{1}{x} \) given \( x = 3 - 2\sqrt{2} \), we proceed as follows:
First, calculate \( \frac{1}{x} \):
\[ x = 3 - 2\sqrt{2} \]
\[ \frac{1}{x} = \frac{1}{3 - 2\sqrt{2}} \]
Rationalize the denominator:
\[ \frac{1}{x} = \frac{1}{3 - 2\sqrt{2}} \cdot \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}} \]
\[ \frac{1}{x} = \frac{3 + 2\sqrt{2}}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} \]
Calculate the denominator:
\[ (3 - 2\sqrt{2})(3 + 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]
So,
\[ \frac{1}{x} = 3 + 2\sqrt{2} \]
Now, find \( x + \frac{1}{x} \):
\[ x + \frac{1}{x} = (3 - 2\sqrt{2}) + (3 + 2\sqrt{2}) \]
Combine like terms:
\[ x + \frac{1}{x} = 3 - 2\sqrt{2} + 3 + 2\sqrt{2} \]
\[ x + \frac{1}{x} = 6 \]
Therefore, \( x + \frac{1}{x} = \boxed{6} \).