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Cho A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)
CMR:\(\frac{7}{12}< A< \frac{5}{6}\)
A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.................+\frac{1}{99.100}\)
CMR: 7/12< A< 5/6
1, CMR: \(\frac{7}{12}<\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}<\frac{5}{6}\)
2, CMR: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{47.48}+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(Cho\) \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(CMR\) \(\frac{7}{12}< A< \frac{5}{6}\)
1, CMR: \(\frac{7}{12}<\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}<\frac{5}{6}\)
2, CMR: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{47.48}+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
Chứng minh rằng \(\frac{7}{12}< A< \frac{5}{6}\)
cho\(A=\frac{1}{1.2}+\frac{1}{3.4}+.........+\frac{1}{99.100}CMR\frac{7}{12}< A< \frac{5}{6}\)
a)Afrac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+...+frac{1}{99.100} 1b)Bfrac{1}{3}+left(frac{1}{3}right)^2+left(frac{1}{3}right)^3+...+left(frac{1}{3}right)^{100} frac{1}{2}c)frac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+...+frac{1}{49.50}frac{1}{26}+frac{1}{27}+...+frac{1}{50}d)Afrac{1}{1.2}+frac{1}{3.4}+...+frac{1}{99.100}.CMRfrac{7}{12} A frac{5}{6}AI ĐÚNG MINK left(TICKright)CHO (làm đc trên 2 câu)
Đọc tiếp
a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)
b)B=\(\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{100}< \frac{1}{2}\)
c)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
d)A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}.CMR\frac{7}{12}< A< \frac{5}{6}\)
AI ĐÚNG MINK \(\left(TICK\right)\)CHO (làm đc trên 2 câu)
Cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)Chứng minh \(\frac{7}{12}\)< A < \(\frac{5}{6}\)
Chứng minh \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)