\(\left|x-2012\right|+\left|x-2013\right|=2014\)
+ TH1: \(x< 2012.\)
\(\Leftrightarrow2012-x+2013-x=2014\)
\(\Rightarrow4025-2x=2014\)
\(\Rightarrow4025-2014=2x\)
\(\Rightarrow2011=2x\)
\(\Rightarrow x=2011:2\)
\(\Rightarrow x=\frac{2011}{2}\left(TM\right).\)
+ TH2: \(2012\le x< 2013.\)
\(\Leftrightarrow x-2012+2013-x=2014\)
\(\Rightarrow1=2014\left(loại\right).\)
+ TH3: \(x\ge2013.\)
\(\Leftrightarrow x-2012+x-2013=2014\)
\(\Rightarrow2x-4025=2014\)
\(\Rightarrow2x=2014+4025\)
\(\Rightarrow2x=6039\)
\(\Rightarrow x=6039:2\)
\(\Rightarrow x=\frac{6039}{2}\left(TM\right).\)
Vậy \(x\in\left\{\frac{2011}{2};\frac{6039}{2}\right\}.\)
Chúc bạn học tốt!
\(\left|x-2012\right|+\left|x-2013\right|=2014\) (1 )
Ta có bảng xét dấu
+) Nếu x < 2012 thì \(\left|x-2012\right|+\left|x-2013\right|\)= ( 2012 - x ) +( 2013 - x) = 2012 - x + 2013 - x = 4025 - 2x
\(\Rightarrow\left(1\right)\Leftrightarrow\) 4025 - 2x = 2014
\(\Leftrightarrow2x=2011\)
\(\Leftrightarrow x=\frac{2011}{2}\) ( thỏa mãn x < 2012)
\(\Rightarrow x=\frac{2011}{2}\) thỏa mãn đề bài
+) Nếu \(2012\le x\le2013\) thì | x - 2012| +|x - 2013 | = ( x - 2012 ) + ( 2013 - x) = x - 2012 + 2013 - x = 1
\(\Rightarrow\left(1\right)\Leftrightarrow\) 1 = 2014 ( vô lí)
+) Nếu x > 2013 thì | x - 2012| + | x - 2013 | = x- 2012 + x - 2013 = 2x - 4025
\(\Rightarrow\left(1\right)\Leftrightarrow\) 2x - 4025=2015
\(\Leftrightarrow2x=2014+4025=6039\)
\(\Leftrightarrow x=\frac{6039}{2}\) ( thỏa mãn x > 1013)
\(\Rightarrow x=\frac{6039}{2}\) thỏa mãn đề bài
Vậy \(x\in\left\{\frac{6039}{2};\frac{2011}{2}\right\}\)
@@ Học tốt @@
## Chiyuki Fujito
