Cho \( \frac{1}{{b + c}} + \frac{1}{{a + c}} + \fr...

A

- Hướng dẫn giải

Ta có:

$\begin{array}{l} \frac{{x - a}}{{b + c}} + \frac{{x - b}}{{a + c}} + \frac{{x - c}}{{a + b}} = 3 \Leftrightarrow \frac{{x - a}}{{b + c}} + \frac{{x - b}}{{a + c}} + \frac{{x - c}}{{a + b}} - 3 = 0\\ \to \left( {\frac{{x - a}}{{b + c}} - 1} \right) + \left( {\frac{{x - b}}{{a + c}} - 1} \right) + \left( {\frac{{x - c}}{{a + b}} - 1} \right) = 0\\ \to \frac{{x - a - b - c}}{{b + c}} + \frac{{x - b - a - c}}{{a + c}} + \frac{{x - c - a - b}}{{a + b}} = 0\\ \to (x - a - b - c)\left( {\frac{1}{{b + c}} + \frac{1}{{a + c}} + \frac{1}{{a + b}}} \right) = 0\\ \to x - a - b - c = 0 \to x = a + b + c \end{array}$