Cho a+b+c=0. Rút gọn biểu thức: 1/(a^2+b^2-c^2)+ 1/(-a^2+b^2+c^2) +1/(a^2-b^2+c^2)
Giải thích
Do a+b+c=0=>a+b=−c=>a2+2ab+b2=c2
Nên a2+b2−c2=−2abb2+c2−a2=−2bca2+c2−b2=−2ac
Vậy 1a2+b2−c2+1b2+c2−a2+1c2+a2−b2
=−12ab+12bc+12ca=−a+b+c2abc
Do a+b+c=0=>a+b=−c=>a2+2ab+b2=c2
Nên a2+b2−c2=−2abb2+c2−a2=−2bca2+c2−b2=−2ac
Vậy 1a2+b2−c2+1b2+c2−a2+1c2+a2−b2
=−12ab+12bc+12ca=−a+b+c2abc