\(A=\sqrt{x-2\sqrt{x-3}}\\ A=\sqrt{x-3-2\sqrt{x-3}+1+2}\\ A=\sqrt{\left(\sqrt{x-3}-1\right)^2+2}\)
vì:
\(\sqrt{x-3}\ge0\\ \Rightarrow\sqrt{x-3}+1\ge1\\ do\:đó\:\left(\sqrt{x-3}+1\right)^2+2\ge2\Rightarrow A\ge\sqrt{2}\)
đẳng thức xảy ra khi x=3
vậy \(MIN_A=\sqrt{2}\) tại x=3
\(B=\sqrt{\left(x-2007\right)^2}+\sqrt{\left(x-1\right)^2}\\ B=\left|x-2007\right|+\left|x-1\right|\\ B=\left|2007-x\right|+\left|x-1\right|\ge\left|2007-x+x-1\right|=2006\)
đẳng thức xảy ra khi :
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2007-x\ge0\\x-1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2007-x< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le2007\\x\ge1\end{matrix}\right.\left(nhận\right)\\\left\{{}\begin{matrix}x>2007\\x< 1\end{matrix}\right.\left(loại\right)\end{matrix}\right.\)
vậy GTNN của B= 2006 tại \(1\le x\le2007\)
