Câu hỏi của haitani anh em

Question

T-07 · Accepted Answer

$x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2019\right)+2019=2019\
=>\left(x+x+x+...+x\right)+\left(1+2+...+2019\right)=0\
=>2020x+\dfrac{\left(2019+1\right).2019}{2}=0\
=>2020x+\dfrac{2020.2019}{2}=0\
=>2020.\left(x+\dfrac{2019}{2}\right)=0\
=>x+\dfrac{2019}{2}=0\
=>x=-\dfrac{2019}{2}\left(KTM\right)$Câu trả lời: $x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2019\right)+2019=2019\
=>\left(x+x+x+...+x\right)+\left(1+2+...+2019\right)=0\
=>2020x+\dfrac{\left(2019+1\right).2019}{2}=0\
=>2020x+\dfrac{2020.2019}{2}=0\
=>2020.\left(x+\dfrac{2019}{2}\right)=0\
=>x+\dfrac{2019}{2}=0\
=>x=-\dfrac{2019}{2}\left(KTM\right)$

OH-YEAH^^ · Answer

$x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2019\right)+2019=2019$$\Rightarrow2020x+\left[\left(2019+1\right).2019:2\right]=0$$\Rightarrow2020x+2020.2019:2=0$$\Rightarrow2020x+2039190=0$$\Rightarrow x=\dfrac{2019}{2}$Câu trả lời: $x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2019\right)+2019=2019$$\Rightarrow2020x+\left[\left(2019+1\right).2019:2\right]=0$$\Rightarrow2020x+2020.2019:2=0$$\Rightarrow2020x+2039190=0$$\Rightarrow x=\dfrac{2019}{2}$

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)