M = |1-x|+|x-2012| >= |1-x+x-2012| = 2011
Dấu "=" xảy ra <=> (1-x) . (x-2012) >= 0
<=> 1 <= x <= 2012
Vậy Max M = 2011 <=> 1 <= x <= 2012
k mk nha
Ta có:
\(\left|x-1\right|\ge x-1\forall x\)
\(\left|x-2012\right|=\left|2012-x\right|\ge2012-x\forall x\)
\(\Rightarrow\left|x-1\right|+\left|x-2012\right|\ge2011\)
Dấu "=" xảy ra khi:
\(\hept{\begin{cases}x-1\ge0\\x-2012\le0\end{cases}}\)
\(\Rightarrow1\le x\le2012\)
M = |1-x|+|x-2012| >= |1-x+x-2012| = 2011
Dấu "=" xảy ra <=> (1-x) . (x-2012) >= 0
<=> 1 <= x <= 2012
Vậy Max M = 2011 <=> 1 <= x <= 2012
\(M=\left|x-1\right|+\left|x-2012\right|\\ =\left|1-x\right|+\left|x-2012\right|\\ \ge\left|1-x+x-2012\right|=2011\)
Dấu "=" xảy ra khi \(\left(1-x\right)\left(x-2012\right)\ge0\\ \Leftrightarrow\hept{\begin{cases}1-x\ge0\\x-2012\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge2012\end{cases}}\)
Min M = 2011 khi x<=1 và x<=2012
