Question

cho \(\left(a+b+c\right)^2=3.\left(a.b+b.c+c.a\right)\) CMR a=b=c

Answer 1

\(3.\left(ab+bc+ca\right)=\left(a+b+c\right)^2\)

\(=>3ab+3bc+3ca=a^2+b^2+c^2+2ab+2bc+2ca\)

\(=>3ab+3bc+3ca-a^2-b^2-c^2-2ab-2bc-2ca=0\)

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

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

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

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

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

Tổng 3 số không âm =0 <=> chúng = 0

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