Question

cho a²+b²+c²=3abc chung minh   a=b=c hoac a+b+c =0

Answer 1

mk chỉnh lại đề nhé: 

cho \(a^3+b^3+c^3=3abc\)

CMR: \(a=b=c\)hoặc   \(a+b+c=0\)

      BÀI LÀM

\(a^3+b^3+c^3=3abc\)

<=>  \(a^3+b^3+c^3-3abc=0\)

<=> \(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

<=> \(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

<=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

<=>  \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)

Xét:  \(a^2+b^2+c^2-ab-bc-ca=0\)

<=>  \(2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

<=>  \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=>  \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)

<=>  \(a=b=c\)

=> đpcm