Rút gọn:
\(\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
Chứng tỏ rằng: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}\)
Chứng tỏ rằng;
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}\)
Tính tổng:
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2019}\)
Tính:F= \(1.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+...+\frac{1}{2019}.\frac{1}{2020}\)
Tìm x biết
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
a)\(\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+......+\frac{1}{5^{2019}}< \frac{1}{2}\)
b) \(\frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^3}+......+\frac{1}{4^2}< 1\)
c) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
d) \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+......+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
e) \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+.....+\frac{1}{79}+\frac{1}{80}>\frac{7}{12}\)
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Thực hiện phép tính (hợp lí nếu có thể):
\(1,\frac{-1}{3}-\frac{-3}{5}-\frac{1}{6}+\frac{1}{43}-\frac{-3}{7}+\frac{-1}{2}-\frac{1}{35}\\ \\ 2,\left(-\frac{1}{3}+\frac{7}{13}\right)-\left(\frac{-16}{24}+\frac{6}{26}+\frac{9}{13}\right)\)\(3,\frac{-7}{3}-\left[\frac{2}{5}-\left(\frac{1}{3}+\frac{-5}{25}\right)\right]\\ 4,\left(2\frac{1}{4}-3\frac{1}{5}\right)-\left[\frac{-3}{4}+\left(\frac{4}{5}-2019\right)\right]\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2019^2}CMR\)\(A< 1\)