x+(x+1)+(x+2)+...+(x+2018)+(x+2019)=2019x+(x+1)+(x+2)+...+(x+2018)+(x+2019)=2019
⇔2020x+(1+2+...+2019)=2019
⇔2020x+(1+2+...+2019)=2019
⇔2020x+(2019+1).2019/2=2019
⇔2020x+(2019+1).2019/2=2019
⇔2020x+2039190=2019
⇔2020x+2039190=2019
⇔x≈−1008
Các câu hỏi tương tự
Tìm x : x+(x+1)+(x+2)+...+2018+2019=2019
tìm x biết : x+(x+1)+(x+2)+...+2018+2019=2019
TÌM X THUỘC Z:
x +(x+1)+(x+2)+...+2018+2019=2019
tìm x ∈ Z
x+(x+1)+(x+2)+...+2018+2019=2019
x +(x+1) + ( x+2)+...+2018+2019=2019
x+(x+1)+(x+2)+....+2018+2019=2019
X+(X+1)+(x+2)+...+2018+2019=2019
x+(x+1)+(x+2)+......+2018+2019=2019
x+(x+1)+(x+2)+...2018+2019=2019