\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
Có: \(\frac{7}{12}=0,58\left(3\right);\frac{99}{100}=0,99;\frac{5}{6}=0,8\left(3\right)\)
Và: \(0,58< 0,99>0,8\left(3\right)\) ( đề sai bạn ơi )