\(X-\frac{2}{3}=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)
\(=>X-\frac{2}{3}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=>X-\frac{2}{3}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>X-\frac{2}{3}=1-\frac{1}{100}\)
\(=>X-\frac{2}{3}=\frac{100}{100}-\frac{1}{100}\)
\(=>X-\frac{2}{3}=\frac{99}{100}\)
\(=>X=\frac{99}{100}+\frac{2}{3}\)
\(=>X=\frac{497}{300}\)
Lưu ý: dấu chấm thay dấu nhân
\(x-\frac{2}{3}=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)
Tổng vế phải gồm : \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
Với vế trái, ta có : \(x-\frac{2}{3}=\frac{99}{100}\)
\(x-\frac{2}{3}=\frac{99}{100}\)
\(x=\frac{99}{100}+\frac{2}{3}\)
\(x=\frac{497}{300}\)
