Cần chứng minh Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(\left(\left|a\right|+\left|b\right|\right)^2\ge\left(\left|a+b\right|\right)^2\)
\(\Leftrightarrow a^2+b^2+2\left|ab\right|\ge a^2+b^2+2ab\)
\(\Leftrightarrow\left|ab\right|\ge ab\) luôn đúng
Dấu = khi \(ab\ge0\)
Áp dụng vào bài ta có:
\(\left|x-2001\right|+\left|x-1\right|\ge\left|x-2001+1-x\right|=2000\)
Dấu = khi \(\left(x-2001\right)\left(x-1\right)\ge0\)\(\Rightarrow1\le x\le2001\)
\(\Rightarrow\begin{cases}\left(x-2001\right)\left(x-1\right)\\1\le x\le2001\end{cases}\)\(\Leftrightarrow\begin{cases}x=1\\x=2001\end{cases}\)
Vậy MinA=2000 khi \(\begin{cases}x=1\\x=2001\end{cases}\)
Ta có
\(A=\left|x-2001\right|+\left|x-1\right|\)
\(\Rightarrow A=\left|2001-x\right|+\left|x-1\right|\)
Vì \(\begin{cases}\left|2001-x\right|\ge2001-x\\\left|x-1\right|\ge x-1\end{cases}\)\(\Rightarrow\left|2001-x\right|+\left|x-1\right|\ge2001-x+x-1\)
\(\Rightarrow\left|2001-x\right|+\left|x-1\right|\ge2000\)
Dấu " = " xảy ra khi \(\begin{cases}2001-x\ge0\\x-1\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\le2001\\x\ge1\end{cases}\)
Vậy MAXA=2000 khi \(1\le x\le2001\)
Ta có :
\(A=\left|x-2001\right|+\left|x-1\right|\\ =>A=\left|2001-x\right|+\left|x-1\right|\\ \)
Vì \(\begin{cases}\left|2001-x\right|\ge2001-x\\\left|x-1\right|\le x-1\end{cases}=>\left|2001-x\right|+\left|x-1\right|\ge2001-x+x+1\)
Dấu = xảy ra khi \(\begin{cases}2001-x\ge0\\x-1\ge0\end{cases}< =>\begin{cases}x\le2001\\x\ge1\end{cases}\)
Vậy MAXA=2000 khi \(1\le x\le2001\)
tại sao |x-2001|=|2001-x|