[tex]\it |x-2019|+|x-m|=m-2019\ \ \ \ (1)\\ \\ \it |x-2019|+|x-m|\geq0 \stackrel{(1)}{\Longrightarrow} m-2019\geq 0 \Rightarrow m\geq2019\ \ \ \ \ (2)\\ \\ I)\ x\leq2019 \stackrel{(2)}{\Longrightarrow} ecua\c{\it t}ia\ devine:\\ \\ -x+2019-x+m=m-2019 \Rightarrow -2x=-2\cdot2019 \Rightarrow x=2019 \Rightarrow S_1=\{2019\}[/tex]
[tex]\it II)\ x\in(2019,m) \stackrel{(2)}{\Longrightarrow} ecua\c{\it t}ia\ devine:\\ \\ x-2019+m-x=m-2019 \Rightarrow 0x=0 \Rightarrow S_2=(2019,\ m)\\ \\ III)\ x\geq m \stackrel{(2)}{\Longrightarrow} ecua\c{\it t}ia\ devine:\\ \\ x-2019+x-m=m-2019 \Rightarrow 2x=2m \Rightarrow x=m \Rightarrow S_3=\{m\}[/tex]
Mulțimea soluțiilor ecuației (1) este :
[tex]\it S=S_1\cup S_2\cup S_3 = [2019,\ m][/tex]
