Question

x+(x+1)+(x+2)+...+199+200=200 tìm x

 

Cho x∈ZSo Sánh So Sánh x²và2x

Answer 1

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+199\right)+\left(x+200\right)=200\)

\(\Leftrightarrow200x+\left(1+2+3+..+200\right)=200\)

\(\Leftrightarrow200x+\dfrac{\left(200+1\right).200}{2}=200\)

\(\Leftrightarrow200x+20100=200\)

\(\Leftrightarrow200x=-19900\)

\(\Leftrightarrow x=-99,5\)

Answer 2

a) Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+199\right)+\left(x+200\right)=200\)

\(\Leftrightarrow x+x+1+x+2+...+x+199+x+200=200\)

\(\Leftrightarrow201x+20100=200\)

\(\Leftrightarrow201x=-19900\)

hay \(x=\dfrac{-19900}{201}\)

Vậy: \(x=\dfrac{-19900}{201}\)