\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{51}+\frac{1}{99}+\frac{1}{100}\)
\(A=\left(\frac{1}{51}+\frac{1}{51}\right)+\frac{1}{52}+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{2}{51}+\frac{1}{52}+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{155}{1652}+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{5999}{87516}+\frac{1}{100}\)
\(A=0.078547465\)
\(CMR:\) \(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{99}-\frac{1}{100}=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
cho A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
chung minh rang
a) A =\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
b) \(\frac{25}{75}+\frac{25}{100}< A< \frac{25}{51}+\frac{25}{75}\)
Cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
CMR:
1, A = \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
2, \(\frac{25}{75}+\frac{25}{100}< A< \frac{25}{51}+\frac{25}{75}\)
CMR :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{100}\)
\(cmr;\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+.....+\frac{1}{99\times100}=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+.....+\frac{1}{100}\)
ai làm đung mình tick cho
Cho \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....+\frac{1}{99\cdot100}\)
\(B=\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+.....+\frac{1}{100}\)
Khi đó A-b=????
Cho A=\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+........+\frac{1}{99\cdot100}\)và B=\(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+.......+\frac{1}{100}\)khi đó A-B =.......? ai giải thích hộ mk vs nhá. thanks
Cho A=\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100},CMR:\frac{3}{5}< A< \frac{31}{40}\)
A=\(\frac{\frac{1}{9}-\frac{1}{6}-\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}:5\frac{1}{6}\)tính A
