\(A=\left|x-a\right|+\left|x-b\right|+\left|x-c\right|+\left|x-d\right|\)
Ta có:
\(\left\{{}\begin{matrix}\left|x-a\right|\ge0\forall x,a\\\left|x-b\right|\ge0\forall x,b\\\left|x-c\right|\ge0\forall x,c\\\left|x-d\right|\ge0\forall x,d\end{matrix}\right.\)
\(\Rightarrow\left|x-a\right|+\left|x-b\right|+\left|x-c\right|+\left|x-d\right|\ge0\) \(\forall x,a,b,c,d.\)
\(\Rightarrow A\ge0.\)
Dấu '' = '' xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-a\right|=0\\\left|x-b\right|=0\\\left|x-c\right|=0\\\left|x-d\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-a=0\\x-b=0\\x-c=0\\x-d=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=a\\x=b\\x=c\\x=d\end{matrix}\right.\Rightarrow a=b=c=d=x.\)
Vậy \(MIN_A=0\) khi \(a=b=c=d=x.\)
Chúc bạn học tốt!
