In a triangle ABC , the tangents of angles A and B are given by A =...

In any triangle $ABC$, $A + B + C = \pi$, which implies $C = \pi - (A + B)$.

Taking the tangent on both sides:
$\tan C = \tan(\pi - (A + B)) = -\tan(A + B)$

Using the tangent addition formula:
$\tan C = -\dfrac{\tan A + \tan B}{1 - \tan A \tan B} = \dfrac{\tan A + \tan B}{\tan A \tan B - 1}$

Substitute the given values $\tan A = x$ and $\tan B = \dfrac{x^2+1}{2x}$:
$\tan A + \tan B = x + \dfrac{x^2+1}{2x} = \dfrac{2x^2 + x^2 + 1}{2x} = \dfrac{3x^2+1}{2x}$

$\tan A \tan B - 1 = x\left(\dfrac{x^2+1}{2x}\right) - 1 = \dfrac{x^2+1}{2} - 1 = \dfrac{x^2-1}{2}$

Now, substitute these into the expression for $\tan C$:
$\tan C = \dfrac{\dfrac{3x^2+1}{2x}}{\dfrac{x^2-1}{2}} = \dfrac{3x^2+1}{x(x^2-1)} = \dfrac{3x^2+1}{x^3-x}$

We are given that $\tan C = \dfrac{13}{6}$. Equating the two expressions:
$\dfrac{3x^2+1}{x^3-x} = \dfrac{13}{6}$

Cross-multiplying yields:
$18x^2 + 6 = 13x^3 - 13x$

Rearranging the terms to form a cubic equation:
$13x^3 - 18x^2 - 13x - 6 = 0$

By trial and error, testing small integer values:
For $x = 2$: $13(2)^3 - 18(2)^2 - 13(2) - 6 = 104 - 72 - 26 - 6 = 0$.

Thus, $x = 2$ is a root. Factoring out $(x-2)$ gives:
$(x-2)(13x^2 + 8x + 3) = 0$

The quadratic factor $13x^2 + 8x + 3 = 0$ has a discriminant $\Delta = 8^2 - 4(13)(3) = 64 - 156 < 0$, so it has no real roots.

Therefore, the only real solution is $x = 2$, which satisfies the condition $x > 1$.

Answer: $2$