Vì a+b+c=0=>(a+b)=-c. Tương tự:(b+c)=-a;(a+c)=-b.
Ta có A=:\(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\frac{b^2}{\left(b-c\right)\left(b+c\right)-a^2}+\frac{c^2}{\left(c-a\right)\left(c+a\right)-b^2}\)
\(=\frac{a^2}{\left(a-b\right).\left(-c\right)-c^2}+tươngtự\)
\(=\frac{a^2}{-ca+bc-c^2}\)+ tương tự
\(=\frac{a^2}{c\left(b-c-a\right)}+tươngtự\)
\(=\frac{a^2}{c\left(b-\left(c+a\right)\right)}\)+ tương tự nha
\(=\frac{a^2}{c\left(b-\left(-b\right)\right)}+tươngtự=\frac{a^2}{2bc}+tươngtự\)
Sau đó ta có :\(\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2bc}\)
=\(\frac{a^3+b^3+c^3}{2abc}=\frac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{2abc}\)
\(=\frac{\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b\right)}{2abc}\)=\(\frac{0-0-3ab\left(-c\right)}{2abc}\)(do a+b+c=0)
=\(\frac{3abc}{2abc}=\frac{3}{2}\)Ok r bạn
