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

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

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

\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow a=b=c}\)

TL:

1) 

Ta có:  \(2a^2+2b^2+2c^2=2ab+2ac+2bc\) 

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

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

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

\(\Rightarrow\left(a-b\right)^2=0\) và\(\left(a-c\right)^2=0\)  và  \(\left(b-c\right)^2=0\) 

\(\Rightarrow a-b=0\) và \(â-c=0\) và  \(b-c=0\) 

=>a=b=c(đpcm)

hình như câu B đề sai bạn nhé!

mk sửa lại ko biết có đúng ko:)

\(â^2+b^2+c^2=2\left(a+b+c\right)\)

hc tốt

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

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=6\left(a^2+b^2+c^2\right)-6\left(ab+bc+ca\right)\)

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

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

\(\Leftrightarrow a=b=c\).

bạn lớp 7 mà học kém quá nhỉ

dễ ot

b,c=1