Question

Chứng minh rằng

Nếu (1-x)²+(x-y)²+(y-z)²=0 thì x=y=z=1

Answer 1

Với mọi giá trị của \(x;y;z\in R\) ta có:

\(\left(1-x\right)^2\ge0;\left(x-y\right)^2\ge0;\left(y-z\right)^2\ge0\)

\(\Rightarrow\left(1-x\right)^2+\left(x-y\right)^2+\left(y-z\right)^2\ge0\)

Để \(\left(1-x\right)^2+\left(x-y\right)^2+\left(y-z\right)^2=0\) thì

\(\left\{{}\begin{matrix}\left(1-x\right)^2=0\\\left(x-y\right)^2=0\\\left(y-z\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}1-x=0\\x-y=0\\y=z=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\1-y=0\\y-z=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\\1-z=0\end{matrix}\right.\Rightarrow x=y=z=1\) (đpcm)

Chúc bạn học tốt!!!

Answer 2

Ta có: \(\left\{{}\begin{matrix}\left(1-x\right)^2\ge0\\\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\end{matrix}\right.\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}1-x=0\\x-y=0\\y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=x\\z=y\end{matrix}\right.\Leftrightarrow x=y=z=1\)

=> đpcm