Các câu hỏi tương tự
So sánh:
\(A=1.3.5.7...97.99\)
\(B=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.\frac{54}{2}...\frac{99}{2}.\frac{100}{2}\)
Chứng minh rằng :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\)
Tính:\(\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\right):\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\right)\)
so sánh:
\(c=1.3.5.7.....99\)và\(D=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.\frac{54}{2}....\frac{100}{2}\)
\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+.......+\frac{1}{98}+\frac{1}{99}+\frac{1}{100}>\frac{1}{2}\)
Chứng minh :
\(\frac{1}{2}< \frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+....+\frac{1}{100}< 1\)
So sánh: 1.3.5....99 với \(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}\)
C= 1.3.5.7..99 với D= \(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
so sánh C=\(\frac{51}{2}.\frac{52}{2}\frac{53}{2}........\frac{100}{2}\)D=1.3.5.7....99