\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}\)
\(=\frac{1}{1.2}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{1.2}+\left(\frac{1}{3}-\frac{1}{100}\right)\)
\(=\frac{1}{1.2}+\frac{97}{300}=\frac{247}{300}\)
\(\text{Vậy }\)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{247}{300}\)
\(\frac{1}{1.2}\)+(\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{99.100}\))
\(\frac{1}{1.2}\)+(\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\))
\(\frac{1}{2}\)+\(\frac{1}{3}-\frac{1}{100}\)
=\(\frac{1}{2}+\frac{97}{300}\)
=\(\frac{247}{300}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
k mik nha
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}\)= \(\frac{99}{100}\)