In a Delta ABC,the sides a,b,c are the roots | vedclass

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(C) Given that $a, b, c$ are the roots of the equation $x^3 - 11x^2 + 38x - 40 = 0$.By Vieta's formulas,we have:$a + b + c = 11$$ab + bc + ca = 38$$ab

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$\frac{9}{16}$

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(C) Given that $a, b, c$ are the roots of the equation $x^3 - 11x^2 + 38x - 40 = 0$.By Vieta's formulas,we have:$a + b + c = 11$$ab + bc + ca = 38$$abc = 40$Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting these into the expression:$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2 + c^2 - a^2}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{2abc}$$= \frac{b^2 + c^2 - a^2 + a^2 + c^2 - b^2 + a^2 + b^2 - c^2}{2abc} = \frac{a^2 + b^2 + c^2}{2abc}$We know $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = (11)^2 - 2(38) = 121 - 76 = 45$.Thus,the expression equals $\frac{45}{2(40)} = \frac{45}{80} = \frac{9}{16}$.

In a $\Delta ABC$,the sides $a$,$b$,$c$ are the roots of the equation $x^3 - 11x^2 + 38x - 40 = 0$; then $\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = $

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