x + (x+1) +(x+2) +......+ (x+2010)=2029099
=> (x+x+....+x) + (1+2+3+.....+2010)=2029099
=> 2011x + 2021055 = 2029099
=> 2011x=8044
=> x = 4
*** k mk nha!
\(x+\left(x+1\right)+\left(x+2\right)+..+\left(x+2020\right)=2029099\)
\(\Leftrightarrow x+x+1+x+2+...+x+2010=2029099\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+..+2010\right)=2029099\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left\{\frac{\left(2010+1\right).\left[\left(2010-1\right):1+1\right]}{2}\right\}=2029099\)
\(\Leftrightarrow2011x+2021055=2029099\)
\(\Leftrightarrow2011x=8044\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
