Với mọi `x` ta luôn có: ` |2x+1| ≥0; |2x+2| ≥0; .....; | 2x+1007| ≥0`
=> ` |2x+1|+|2x+2|+|2x+3|+...+|2x+1007|` `≥0`
=> `2015x ≥0`
=> `x ≥0`
Vì x ≥0 nên: `|2x+1| = 2x+1`; `|2x+2| = 2x+2`;.... ; `| 2x+1007| = 2x+1007`
=> `2x + 1 + 2x + 2 + 2x+3+...+ 2x+1007 = 2015x`
=> `2x + 2x + 2x+...+ 2x + 1+2+3+...+1007 = 2015x`
Đặt `A = 1+2+3+...+1007`
Tổng A là: `[(1007- 1):1 +1]. ( 1007+1):2= 507528`
=> `2x.1007 + 507528 = 2015x`
=> `2014x + 507528 = 2015x`
=> `x = 507528`
Vậy `x = 507528`
