Có công thức : l a l + l b l > l a + b l

=> l x- 2 l + l x - 3 l 

=> l x - 2 l + l 3 - x l 

=>  l x - 2 l + l 3 - x l  \(\ge\)l x - 2 + 3 - x l ( = 1 )

=> Vậy GTNN là 1 khi x \(\ge\)3

=> 

Ta có: |x - 2| + |x - 3| = |x - 2| + |3 - x| > = |x - 2 + 3 - x| = |1| = 1

Dấu "=" xảy ra<=> (x -- 2)(3 - -x ) > = 0 <=> 2 < = x < = 3

Vậy min|x - 2| + |x - 3| = 1 <=> 2 < = x < = 3

Sửa đề : \(\frac{x+3}{2010}+\frac{x+2}{2011}-\frac{x+1}{2012}=\frac{x-1}{2014}\)

\(\Leftrightarrow\frac{x+3}{2010}+1+\frac{x+2}{2011}+1-\frac{x+1}{2012}+1=\frac{x-1}{2014}+1\)

\(\Leftrightarrow\frac{x+3+2010}{2010}+\frac{x+2+2011}{2011}-\frac{x+1+2012}{2012}=\frac{x-1+2014}{2014}\)

\(\Leftrightarrow\frac{x+2013}{2010}+\frac{x+2013}{2011}-\frac{x+2013}{2012}=\frac{x+2013}{1014}\)

\(\Leftrightarrow\frac{x+2013}{2010}+\frac{x+2013}{2011}-\frac{x+2013}{2012}-\frac{x+2013}{2014}=0\)

\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2014}\right)=0\)

\(\Rightarrow x+2013=0\)( Vì (  \(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2014}\)) \(\ne0\))

\(\Rightarrow x=-2013\)

20092009^10 lớn hơn