Let [tex]\( a = \frac{3+\sqrt{2}}{3-\sqrt{2}} \)[/tex] and [tex]\( b = \frac{3-\sqrt{2}}{3+\sqrt{2}} \)[/tex].
To find the sum [tex]\( a + b \)[/tex], we first rationalize the denominators of the fractions for both [tex]\( a \)[/tex] and [tex]\( b \)[/tex].
### Rationalizing [tex]\( a \)[/tex]:
[tex]\[ a = \frac{3 + \sqrt{2}}{3 - \sqrt{2}} \][/tex]
Multiply the numerator and the denominator by the conjugate of the denominator, [tex]\( 3 + \sqrt{2} \)[/tex]:
[tex]\[ a = \frac{(3 + \sqrt{2})(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})} \][/tex]
Simplify the numerator and the denominator:
[tex]\[ = \frac{9 + 6\sqrt{2} + 2}{9 - 2} \][/tex]
[tex]\[ = \frac{11 + 6\sqrt{2}}{7} \][/tex]
### Rationalizing [tex]\( b \)[/tex]:
[tex]\[ b = \frac{3 - \sqrt{2}}{3 + \sqrt{2}} \][/tex]
Similarly, multiply the numerator and the denominator by the conjugate of the denominator, [tex]\( 3 - \sqrt{2} \)[/tex]:
[tex]\[ b = \frac{(3 - \sqrt{2})(3 - \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})} \][/tex]
Simplify the numerator and the denominator:
[tex]\[ = \frac{9 - 6\sqrt{2} + 2}{9 - 2} \][/tex]
[tex]\[ = \frac{11 - 6\sqrt{2}}{7} \][/tex]
### Finding [tex]\( a + b \)[/tex]:
[tex]\[ a + b = \frac{11 + 6\sqrt{2}}{7} + \frac{11 - 6\sqrt{2}}{7} \][/tex]
Combine the fractions:
[tex]\[ = \frac{(11 + 6\sqrt{2}) + (11 - 6\sqrt{2})}{7} \][/tex]
Simplify the expression inside the numerator:
[tex]\[ = \frac{11 + 11}{7} \][/tex]
[tex]\[ = \frac{22}{7} \][/tex]
Thus,
[tex]\[ a + b = \frac{22}{7} \][/tex]
The value of [tex]\( \frac{22}{7} \)[/tex] is approximately [tex]\( 3.142857142857143 \)[/tex].
Therefore, the value of [tex]\( a + b \)[/tex] is:
[tex]\[ \boxed{3.142857142857143} \][/tex]
