Ta có: A = \(\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}\right)\)
Nhận xét: \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};...;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow A>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)=\frac{1}{10}+\frac{90}{100}=1\)
Vậy A > 1 (đpcm)
Cho tổng A =\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ A > 1
Cho tổng A =\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ A > 1
Cho \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ rằng A > 1
Cho tổng A= \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ A>1
cho A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\) chứng tỏ A>1
Cho tổng A = \(\frac{1}{10}\)+ \(\frac{1}{11}\)+ \(\frac{1}{12}\)+ ...... + \(\frac{1}{99}\)+ \(\frac{1}{100}\)
Chứng tỏ rằng A > 1.
Cho \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ rằng \(A>1\)
Chứng tỏ tổng sau lớn hơn 1
\(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Cho \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ A > 1
