In a triangle ABC , the sides $a, \mathrm{~b}, \mathrm{c}$ are such that they are the roots of the equation $x^3-11 x^2+38 x-40=0$ Then $\frac{\cos \mathrm{A}}{a}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}=$

The sides of the triangle are the roots of the equation $x^3 - 11x^2 + 38x - 40 = 0$. Testing integer divisors of 40 reveals that $x = 2$ is a root. Dividing the cubic polynomial by $(x - 2)$ yields $x^2 - 9x + 20$, which factors as $(x - 4)(x - 5)$. The sides are therefore $2$, $4$, and $5$ in some order.

Using the Law of Cosines, $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$, with similar expressions for $\cos B$ and $\cos C$. Substituting into $\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c}$ gives $\frac{b^2 + c^2 - a^2}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{2abc}$.

The sum simplifies to $\frac{a^2 + b^2 + c^2}{2abc}$, as the numerator reduces to $a^2 + b^2 + c^2$ after cancellation.

With side lengths $2$, $4$, and $5$, $a^2 + b^2 + c^2 = 4 + 16 + 25 = 45$ and $abc = 40$. The expression evaluates to $\frac{45}{80} = \frac{9}{16}$.

Final answer: $\frac{9}{16}$