Question

vì a+b+c=0 nên a=-(b+c)\Rightarrow $a^2$=$(b+c)^2$
tương tự ta có : $b^2$=$(a+c)^2$
$c^2$=$(a+b)^2$
\Rightarrow $\frac{a^2}{a^2-b^2-c^2}$+$\frac{b^2}{b^2-c^2-a^2}$+$\frac{c^2}{c^2-b^2-a^2}$
=$\frac{a^2}{(b+c)^2-b^2-c^2}$+$\frac{b^2}{(a+c)^2-a^2-c^2}$
+$\frac{c^2}{(a+b)^2-a^2-b^2}$
=$\frac{a^2}{2bc}$+$\frac{b^2}{2ac}$+$\frac{c^2}{2ab}$
=$\frac{a^3+b^3+c^3}{2abc}$
vì a+b+c=0 nên a^3+b^3+c^3=3abc(hằng đẳng thức nâng cao)
\Rightarrow $\frac{a^3+b^3+c^3}{2abc}$=$\frac{3}{2}$