Question

Cho \(a+b+c=0.\)

Chứng minh rằng :

 \(a^4+b^4+c^4=2\left(ab+bc+ac\right).\)

Answer 1

\(Ta\)\(có\):\(\)

\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\)
\(=\left(a^2+b^2+c^2\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)\(Mà\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0=a^2+b^2+c^2+2\left(ab+ac+bc\right)\)\(=1+2\left(ab+ac+bc\right)=0\Rightarrow2\left(ab+ac+bc\right)=-1\)\(\Rightarrow a^2+b^2+c^2=2\left(ab+bc+ac\right)\)

Answer 2

Xl nha dòng cuối mik ghi nhầm

Phài là \(a^4+b^4+c^4=2\left(ab+bc+ac\right)\)