cute thế bn ơi tick nha mk ko bít lm mk hok loqps 5
Ta có: |x-2015|+|2016-x|>=|x-2015+2016-x|=1(theo công thức : |A|+|B|>=|A+B|
=>đpcm
Đặt \(\left|x-2015\right|+\left|2016-x\right|\)
Nếu \(x<2015\) thì:
\(A=2015-x+x-2016=4031-2x\)
Do \(x<2015\Rightarrow-2x>-4032\)
Nếu \(x>2016\) thì
\(A=x-2015+x-2016=2x-4031\)
Do \(x>2016\Leftrightarrow2x>4032n\text{ê}n\text{A}>1\)
Nếu \(2015\le x\le2016\) thì
\(A=x-2015+2016-x=1\)
Vậy: \(A=\left|x-2015\right|+\left|2016-x\right|\ge1\)
xét x<2015:
=>M=2016-x-(x-2015)
=2016-x-x+2015
=2016+2015-2x>2016+2015-2015.2=1
xét 2015<(=)x<(=)2016:
=>M=x-2015+2016-x=1
xét x>2016:
=>M=x-2015-(2016-x)
=x-2015-2016+x=2x-2015-2016>2016.2-2015.2016=1
=>lx-2015l+l2016-xl>(=)1
=>đpcm