Question

 biết x+(x+1)+(x+2)+...+(x+2016)=2018.2017

tìm x (x la số tự nhiên)

Answer 1

x+(x+1)+(x+2)+...+(x+2016)=2018.2017

x+x+1+x+2+...+x+2016=2018.2017

(x+x+x+...+x)+(1+2+...+2016)=2018.2017

x.2017+{[(2016-1):1+1].(2016+1):2}=2018.2017

x.2017+1008.2017=2018.2017

x.2017=2018.2017-1008.2017

x.2017=2017(2018-1008)

x.2017=2017.1010

=>x=1010

Answer 2

mình cho các bạn biết x=1010 nhé

Answer 3

lm dung co k khong

Answer 4

 x+(x+1)+(x+2)+...+(x+2016)=2018.2017

 x+(x+1)+(x+2)+...+(x+2016)=4070306

(x+x+...+x)+(1+2+3+...+2016)=4070306

2017x+2033136=4070306

2017x=2037170

x=1010