Question

x+(x+1) + (x+2)+....+(x+2010)=2029099

Answer 1

Giải:

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2010\right)=2029099\)

\(\Leftrightarrow x+x+1+x+2+...+x+2010=2029099\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+2010\right)=2029099\)

\(\Leftrightarrow2011x+\dfrac{\left(2010+1\right).2010}{2}=2029099\)

\(\Leftrightarrow2011x+2021055=2029099\)

\(\Leftrightarrow2011x=2029099-2021055\)

\(\Leftrightarrow2011x=8044\)

\(\Leftrightarrow x=\dfrac{8044}{2011}=4\)

Vậy \(x=4\).

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