Cho tam giác ABC có BC = a, CA = b, AB = c thỏa mãn \[\frac{{a + b}}{6} = \frac{{b + c}}{5} = \frac{{c + a}}{7}\]. Tính giá trị của biểu thức T = cosA + 2cosB + 3cosC.

A. \(\frac{{57}}{{16}}\).

B. \(\frac{{16}}{{57}}\).

C. \( - \frac{{57}}{{16}}\).

D. \( - \frac{{16}}{{57}}\).

Đáp án đúng là: A

Ta có \(\left\{ \begin{array}{l}\frac{{a + b}}{6} = \frac{{b + c}}{5}\\\frac{{a + b}}{6} = \frac{{c + a}}{7}\end{array} \right.\)\( \Rightarrow \left\{ \begin{array}{l}5a = b + 6c\\a + 7b = 6c\end{array} \right.\)

\( \Rightarrow \left\{ \begin{array}{l}5a = b + a + 7b\\a + 7b = 6c\end{array} \right.\)\( \Rightarrow \left\{ \begin{array}{l}b = \frac{a}{2}\\a + 7.\frac{a}{2} = 6c\end{array} \right.\)\( \Rightarrow \left\{ \begin{array}{l}b = \frac{a}{2}\\c = \frac{{3a}}{4}\end{array} \right.\)

Khi đó ta có:

⦁ \(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{{{\left( {\frac{a}{2}} \right)}^2} + {{\left( {\frac{{3a}}{4}} \right)}^2} - {a^2}}}{{2.\frac{a}{2}.\frac{{3a}}{4}}} = - \frac{1}{4}\).

⦁ \(\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} = \frac{{{a^2} + {{\left( {\frac{{3a}}{4}} \right)}^2} - {{\left( {\frac{a}{2}} \right)}^2}}}{{2a.\frac{{3a}}{4}}} = \frac{7}{8}\).

⦁ \(\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = \frac{{{a^2} + {{\left( {\frac{a}{2}} \right)}^2} - {{\left( {\frac{{3a}}{4}} \right)}^2}}}{{2a.\frac{a}{2}}} = \frac{{11}}{{16}}\).

Vì vậy \(T = \cos A + 2\cos B + 3\cos C = - \frac{1}{4} + 2.\frac{7}{8} + 3.\frac{{11}}{{16}} = \frac{{57}}{{16}}\).

Vậy ta chọn phương án A.