Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|x-2007\right|=\left|x-1\right|+\left|2007-x\right|\ge\left|x-1+2007-x\right|=2006\)Dấu "=" xảy ra khi \(\left(x-1\right)\left(2007-x\right)\ge0\Rightarrow1\le x\le2007\)
Lại có
\(\left\{\begin{matrix}\left|x-30\right|\ge0\\\left|y-4\right|\ge0\\\left|z-1975\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-1\right|+\left|x-30\right|+\left|y-4\right|+\left|z-1975\right|+\left|x-2007\right|\ge2006\)Dấu "="xảy ra khi \(\left\{\begin{matrix}\left|x-30\right|=0\\\left|y-4\right|=0\\\left|z-1975\right|=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left\{\begin{matrix}x-30=0\\y-4=0\\z-1975=0\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=30\\y=4\\z=1975\end{matrix}\right.\)
Vậy x=30,y=4,z=1975
