Question

tim x, biết: x+( x+1)+(x+2)+...+(x+2003)= 2004

Answer 1

x + (x + 1) + (x + 2) + ... + (x + 2003) = 2004

=> x + x + 1 + x + 2 + .... + x + 2003 = 2004

=> 2004x + 2007006 = 2004

=> 2004x = 2005002

=> x = 1000,5

Answer 2

Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2003\right)=2004\)

\(\Leftrightarrow2004x+2007006=2004\)

\(\Leftrightarrow2004x=-2005002\)

hay \(x=-\dfrac{2001}{2}\)