\(x-\frac{1}{2}-\frac{1}{6}-...-\frac{1}{9900}=200\)
\(\Leftrightarrow x-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)=200\)
\(\Leftrightarrow x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=200\)
\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=200\)
\(\Leftrightarrow x-\left(1-\frac{1}{100}\right)=200\)
Ez rồi :) Tự giải tiếp
Ta có: \(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{9900}=200\)
=> \(x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)=200\)
=> \(x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=200\)
=> \(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)=200\)
=> \(x-\left(1-\frac{1}{100}\right)=200\)
=> \(x-\frac{99}{100}=200\)
=> \(x=200+\frac{99}{100}\)
=> \(x=\frac{20099}{100}\)