Các câu hỏi tương tự
CM: A=\(\frac{1}{2}\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{9240}\right)>\frac{57}{462}\)
Tính:
\(\left(\frac{\left(6-4\frac{1}{2}\right):0,03}{\left(3\frac{1}{20}-2,65\right).4+\frac{2}{5}}-\frac{\left(0,3-\frac{3}{20}\right).1\frac{1}{2}}{\left(1,88+2\frac{3}{25}\right).\frac{1}{80}}\right):\frac{49}{60}\)
Bài 1 :Chứng tỏ rằng :
a) \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
b) \(3< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
CTR\(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}< \frac{3}{2}\)
\(\frac{99}{100}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}< \frac{99}{202}CM\)
CM \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)<\(\frac{3}{4}\)
A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)CM \(\frac{7}{12}< A< \frac{5}{6}\)
KN
2 tháng 5 2019 lúc 21:23
\(Cm:\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
Help me! :((
Chứng minh rằng: \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
11 tháng 8 2020 lúc 15:40
Chứng minh: \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}< \frac{3}{2}\).