Câu hỏi của Nguyễn Thị Thơm

Question

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

Ngô Thanh Sang · Accepted Answer

$x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2003\right)=2004$

$2004x+1+2+3+...+2003=2004$

$2004x+\dfrac{2003.\left(1+2003\right)}{2}=2004$

$2004x=2004-2007006$

$2004x=-2005002$

$x=-2005002:2004=-1000,5=-1000\dfrac{1}{2}$

Vậy:...Câu trả lời: $x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2003\right)=2004$

$2004x+1+2+3+...+2003=2004$

$2004x+\dfrac{2003.\left(1+2003\right)}{2}=2004$

$2004x=2004-2007006$

$2004x=-2005002$

$x=-2005002:2004=-1000,5=-1000\dfrac{1}{2}$

Vậy:...

DƯƠNG PHAN KHÁNH DƯƠNG · Answer

$x+\left(x+1\right)+\left(x+2\right)+\left(x+2013\right)=2014$

$\Rightarrow2004x=\left(1+2+......+2013\right)$ $=2004$

$\Rightarrow2004x=\dfrac{\left(1+2003\right).2003}{2}=2004$

$\Rightarrow2004x=2007006=2004$

$\Rightarrow2004x=2004-2007006$

$\Rightarrow2004x=-2005002$

$\Rightarrow x=-2005002:2004$

$\Rightarrow x=-\dfrac{2001}{2}=-1000,5$Câu trả lời: $x+\left(x+1\right)+\left(x+2\right)+\left(x+2013\right)=2014$