Từ giả thiết ta có:
\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^2=a^2\)
\(\Rightarrow b^2+2bc+c^2=a^2\Rightarrow a^2-b^2-c^2=2bc\)
Tương tự:
\(b^2-c^2-a^2=2ca,c^2-a^2-b^2=2ab\)
Từ đây suy ra:
\(A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{2abc}\)
Mặt khác lại có:
\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^3=-a^3\)
\(\Rightarrow b^3+c^3+3bc\left(b+c\right)=-a^3\Rightarrow a^3+b^3+c^3=-3bc\left(b+c\right)\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
\(\Rightarrow A=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3}{2}\)
Ta có: a+b+c=0
<=>a=-b-c
<=>\(a^2=(-b-c)^2\)
<=>\(a^2=b^2+c^2+2bc\)
cmtt:\(b^2=a^2+c^2+2ac\)
\(c^2=a^2+b^2+2ab\)
=> A=\(\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3}{2abc}+\frac{b^3}{ 2abc}+\frac{c^3}{2abc} \)
=\(\frac{1}{2abc}(a^3+b^3+c^ 3)\)
Cm đẳng thức phụ
Với a+b+c=0=> \(a^3+b^3+c^3=3abc\)
=>A=\(\frac{3}{2} \)