ta có: a+b+c=0
\(\left(a+b\right)^2=\left(-c\right)^2\Rightarrow a^2+2ab+b^2=c^2\Rightarrow a^2+b^2-c^2=-2ab\\ \)
tương tự \(\left(b+c\right)^2=\left(-a\right)^2\Rightarrow b^2+2bc+c^2=a^2\Rightarrow b^2+c^2-a^2=-2bc\\ \)
\(\left(a+c\right)^2=\left(-b\right)^2\Rightarrow a^2+2ac+c^2=b^2\Rightarrow a^2+c^2-b^2=-2ac\\ \)
thế vào biểu thức \(B=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ac}{a^2+c^2-b^2}\)
\(B=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ac}{-2ac}=\frac{-1}{2}+\frac{-1}{2}+\frac{-1}{2}=\frac{-3}{2}\)
ta có a+b+c=0 => a+b= - c => a^2+2ab+b^2=c^2 => a^2+b^2=c^2-2ab
=>b+c= - a => b^2+2bc+c^2=a^2 => b^2+c^2=a^2-2bc
=>a+c= - b => a^2+2ac+c^2=b^2 =>a^2+c^2=b^2-2ac
từ đó ta có: B = ab / (a^2+b^2-c^2) + bc/ (b^2+c^2-a^2) + ca/ (c^2+a^2-b^2)
=> ab/(c^2 -2ab+c^2)+bc/ (a^2 -2bc+a^2) + ca/ (b^2 -2ac+b^2)
=> B= -ab/2ab - bc/2bc - ca/2ca
=> B = -3/2
