Đáp án:
\[B = 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x = \frac{a}{{b + c}}\\
y = \frac{b}{{c + a}}\\
z = \frac{c}{{a + b}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
xb + xc = a\\
yc + ya = b\\
za + zc = c
\end{array} \right.\\
\Rightarrow \left( {xb + xc} \right) + \left( {yc + ya} \right) + \left( {za + zc} \right) = a + b + c\\
\Rightarrow a\left( {y + z - 1} \right) + b\left( {x + z - 1} \right) + c\left( {x + y - 1} \right) = 0,\,\,\,\,\forall a,b,c\\
\Rightarrow \left\{ \begin{array}{l}
y + z - 1 = 0\\
x + z - 1 = 0\\
x + y - 1 = 0
\end{array} \right. \Leftrightarrow x = y = z = \frac{1}{2}\\
\Rightarrow B = xy + yz + zx + 2xyz\\
= \frac{1}{2}.\frac{1}{2} + \frac{1}{2}.\frac{1}{2} + \frac{1}{2}.\frac{1}{2} + 2.\frac{1}{2}.\frac{1}{2}.\frac{1}{2} = 1
\end{array}\)
