Answer :
Answer:
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Step-by-step explanation:
Given that \( x - \sqrt{2} = 3 \), we need to find whether \( \frac{x + 1}{x} \) is rational or irrational.
First, solve for \( x \):
\[ x = 3 + \sqrt{2} \]
Now, calculate \( \frac{x + 1}{x} \):
\[ \frac{x + 1}{x} = \frac{(3 + \sqrt{2}) + 1}{3 + \sqrt{2}} = \frac{4 + \sqrt{2}}{3 + \sqrt{2}} \]
To determine if \( \frac{4 + \sqrt{2}}{3 + \sqrt{2}} \) is rational or irrational, let's rationalize the denominator:
\[ \frac{4 + \sqrt{2}}{3 + \sqrt{2}} \cdot \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{(4 + \sqrt{2})(3 - \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})} \]
Calculate the denominator:
\[ (3 + \sqrt{2})(3 - \sqrt{2}) = 3^2 - (\sqrt{2})^2 = 9 - 2 = 7 \]
Calculate the numerator:
\[ (4 + \sqrt{2})(3 - \sqrt{2}) = 4 \cdot 3 + 4 \cdot (-\sqrt{2}) + \sqrt{2} \cdot 3 + \sqrt{2} \cdot (-\sqrt{2}) \]
\[ = 12 - 4\sqrt{2} + 3\sqrt{2} - 2 \]
\[ = 10 - \sqrt{2} \]
So,
\[ \frac{4 + \sqrt{2}}{3 + \sqrt{2}} = \frac{10 - \sqrt{2}}{7} \]
Since \( \frac{10 - \sqrt{2}}{7} \) is not in the form of \( \frac{p}{q} \) where \( p \) and \( q \) are integers (because it contains \( \sqrt{2} \)), \( \frac{x + 1}{x} \) is irrational.
Therefore, \( \frac{x + 1}{x} \) is **irrational**.