\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2003\right)=2004\)
\(2004x+1+2+3+...+2003=2004\)
\(2004x+\dfrac{2003.\left(1+2003\right)}{2}=2004\)
\(2004x=2004-2007006\)
\(2004x=-2005002\)
\(x=-2005002:2004=-1000,5=-1000\dfrac{1}{2}\)
Vậy:...
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+2013\right)=2014\)
\(\Rightarrow2004x=\left(1+2+......+2013\right)\) \(=2004\)
\(\Rightarrow2004x=\dfrac{\left(1+2003\right).2003}{2}=2004\)
\(\Rightarrow2004x=2007006=2004\)
\(\Rightarrow2004x=2004-2007006\)
\(\Rightarrow2004x=-2005002\)
\(\Rightarrow x=-2005002:2004\)
\(\Rightarrow x=-\dfrac{2001}{2}=-1000,5\)
