Question

Cho a; b; c thỏa mãn:

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

Chứng Minh rằng \(a=b=c\)

 

Answer 1

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

\(\Leftrightarrow\hept{\begin{cases}a+b-2c=a-b\\b+c-2a=b-c\\c+a-2b=c-a\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2b-2c=0\\2c-2a=0\\2a-2b=0\end{cases}}\)

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

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

\(\Leftrightarrow a=b=c\)( đpcm )

Answer 2

\(\Rightarrow\hept{\begin{cases}a+b-2c=a-b\\b+c-2a=b-c\\c+a-2b=a-c\end{cases}}\)

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

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