Question

1. Tìm x

\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

        

Answer 1

\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2019}{2020}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)

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Answer 2

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2020}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)

Rồi bn tự tìm x nha!hok tot

Answer 3

\(\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

\(\Leftrightarrow2\left[\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{x\left(x+1\right)}\right]=\frac{2019}{2020}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{2020}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)\(\Leftrightarrow x+1=4040\)\(\Leftrightarrow x=4039\)

Vậy \(x=4039\)