If $$ x=7+4 \sqrt{3} $$ then find the value of $$ x+\frac{1}{x} $$

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1. Given $$ x = 7 + 4\sqrt{3} $$, we first compute the reciprocal $$ \frac{1}{x} $$.

2. To simplify $$ \frac{1}{7 + 4\sqrt{3}} $$, multiply the numerator and denominator by the conjugate of the denominator:
   $$
   \frac{1}{7+4\sqrt{3}} = \frac{7-4\sqrt{3}}{(7+4\sqrt{3})(7-4\sqrt{3})}.
   $$

3. Evaluate the denominator:
   $$
   (7+4\sqrt{3})(7-4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 16 \times 3 = 49 - 48 = 1.
   $$
   So, 
   $$
   \frac{1}{7+4\sqrt{3}} = 7-4\sqrt{3}.
   $$

4. Now, add $$ x $$ and $$ \frac{1}{x} $$:
   $$
   x+\frac{1}{x} = (7+4\sqrt{3}) + (7-4\sqrt{3}) = 7+7+4\sqrt{3}-4\sqrt{3}.
   $$

5. Simplify the expression:
   $$
   x+\frac{1}{x} = 14.
   $$

Thus, the value of $$ x+\frac{1}{x} $$ is $$ 14 $$.
$$ x + \frac{1}{x} = 14 $$

If then find the value of

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