Question

Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\) =\(\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(c+a-2b\right)^2\) .CMR:a=b=c

Answer 1

Không mất tính tổng quát ta giả sử \(a\ge b\ge c\)

Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\a-c=z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}z\ge x\ge0\\z\ge y\ge0\end{matrix}\right.\)

Ta có:

\(x^2+y^2+z^2=\left(x-y\right)^2+\left(x+z\right)^2+\left(y+z\right)^2\)

\(\Leftrightarrow x^2+y^2+z^2+2xz+2yz-2xy=0\)

\(\Leftrightarrow z^2+2xz+2yz+\left(x-y\right)^2=0\)

Vì \(\Rightarrow\left\{{}\begin{matrix}z\ge x\ge0\\z\ge y\ge0\end{matrix}\right.\)

\(\Rightarrow z^2+2xz+2yz+\left(x-y\right)^2\ge0\)

Dấu = xảy ra khi \(x=y=z=0\)

Hay \(a=b=c\)

Answer 2

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

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-4ab-4bc-4ca\)

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

\(=\left(a+b\right)^2-4\left(a+b\right)c+4c^2+\left(b+c\right)^2-4\left(b+c\right)a+4a^2+\left(a+c\right)^2-4\left(a+c\right)b+4b^2\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-4\left(a+b\right)c+4c^2-4\left(b+c\right)a+4a^2-4\left(a+c\right)b+4b^2\)

Nhìn vào thấy 2 vế có \(\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\) rút gọn luôn thì được

\(-4ab-4bc-4ca=-4\left(a+b\right)c+4c^2-4\left(b+c\right)a+4a^2-4\left(a+c\right)b+4b^2\)

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

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

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

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

Xảy ra khi \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\)

Answer 3

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

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

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

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

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

<=> \(2a^{^2}+2b^2+2c^2-2ab-2ac-2bc=0\)

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

<=> a = b= c ( đpcm )