\(\left(a+b+c\right)^2=a^2+b^2+2\left(ab+bc+ac\right)\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(ab+bc+ac\right)\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)
\(\Rightarrow a=b=c\left(đpcm\right)\)
có\(\left(a^2-2a+1\right).\)\(\left(b^2-2b+1\right)\)\(.\)\(\left(c^2-2c+1\right)\)\(=0\)
\(\left(a-1\right)^2\)\(+\)\(\left(b-1\right)^2\)\(+\left(c-1\right)^2\)
\(Vì\)\(\left(a-1\right)^2;\)\(\left(b-1\right)^2;\)\(\left(c-1\right)^2\ge0\)
Dấu \(=\)xảy ra khi
\(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}a=1\\b=1\\c=1\end{cases}}\)
\(\Rightarrow\)\(a=b=c=1\left(đpcm\right)\)
(a+b+c)2=3(ab+bc+ca)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow a=b=c}\)(đpcm)
\(\left(a+b+c\right)^2=a^2+b^2+2\left(ab+bc+ac\right)\)\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(ab+bc+ac\right)\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)
\(\Rightarrow a=b=c\left(đpcm\right)\)
~ Rất vui vì giúp đc bn ~
