Cho \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{99}+\frac{1}{100}\)SO SÁNH A VỚI 1
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![](https://rs.olm.vn/images/avt/0.png?1311)
A= 1/10+1/11+1/12+1/13+...........+1/99+1/100
2A=1/9+1/10+1/11+1/12+...........+1/98+1/99
2A-A=(1/10+1/11+1/12+1/13+.............+1/99+1/100)-(1/9+1/10+1/11+1/12+............1/98+1/99)
A=1/100-1/9
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{100}\)
\(A< \frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{100.101}\)
\(A< \frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{101}\)
\(A< \frac{1}{10}-\frac{1}{101}=\frac{101}{1010}-\frac{10}{1010}=\frac{91}{1010}< \frac{505}{1010}\)
\(A< \frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)\(+\)\(\frac{1}{10}\)\(-\)\(\frac{1}{11}\)\(+\)\(\frac{1}{11}\)\(-\)\(\frac{1}{12}\)\(+\)\(\frac{1}{12}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{14}\)\(+\)\(\frac{1}{14}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{2}{45}\)
\(A=\left(\frac{1}{9}.\frac{1}{10}+\frac{1}{10}.\frac{1}{11}\right)+\left(\frac{1}{11}.\frac{1}{12}+\frac{1}{12}.\frac{1}{13}\right)+\left(\frac{1}{13}.\frac{1}{14}+\frac{1}{14}.\frac{1}{15}\right)\)
Sau đó nhân phân phối ra là xong nhé bạn
A><=1
k mình nha