chứng minh
\(\dfrac{1}{1975^2}+\dfrac{1}{1976^2}+\dfrac{1}{1977^2}+...+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}< \dfrac{1}{1974}\)
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KJ
1
8 tháng 1 2022
\(20M=\dfrac{20^{1976}+1+19}{20^{1976}+1}=1+\dfrac{19}{20^{1976}+1}\)
\(20N=\dfrac{20^{1977}+1+19}{20^{1977}+1}=1+\dfrac{19}{20^{1977}+1}\)
mà \(20^{1976}+1< 20^{1977}+1\)
nên M>N
3 tháng 11 2018
Ta có : P = \(\dfrac{1}{1975}\left(\dfrac{2}{1945}-1\right)-\dfrac{1}{1945}\left(1-\dfrac{2}{1975}\right)+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)
\(-\dfrac{3}{1975.1945}\)
= \(\dfrac{2}{1975.1945}-\dfrac{1}{1975}-\dfrac{1}{1945}+\dfrac{2}{1975.1945}+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)
\(-\dfrac{3}{1975.1945}\)
= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)
= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)
= \(\dfrac{1973}{1975}\)
9 tháng 9 2017
\(\frac{1}{1975^2}+\frac{1}{1976^2}+...+\frac{1}{2017^2}< \frac{1}{1974.1975}+\frac{1}{1975.1976}+...+\frac{1}{2016.2017}\)
\(=\frac{1}{1974}-\frac{1}{1975}+\frac{1}{1975}-\frac{1}{1976}+...+\frac{1}{2016}-\frac{1}{2017}=\frac{1}{1974}-\frac{1}{2017}< \frac{1}{1974}\)
18 tháng 3 2018
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
MN
1
MV
5 tháng 5 2017
\(A=\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{2016}{2^{2017}}\\ 2A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2016}{2^{2016}}\\ 2A-A=\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2016}{2^{2016}}\right)-\left(\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{2016}{2^{2017}}\right)\\ A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{2016}{2^{2017}}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2016}{2^{2016}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2016}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{2016}{2^{2017}}\right)\\ A=1-\dfrac{2017}{2^{2016}}-\dfrac{2016}{2^{2017}}\\ A=1-\dfrac{4034}{2^{2017}}-\dfrac{2016}{2^{2017}}\\ A=1-\left(\dfrac{4034}{2^{2017}}+\dfrac{2016}{2^{2017}}\right)\\ A=1-\dfrac{6050}{2^{2017}}< 1\)
Vậy \(A< 1\)
Ta có : \(\dfrac{1}{1794}\)>\(\dfrac{1}{1795^2}\)
\(\dfrac{1}{1794}\)>\(\dfrac{1}{1796^2}\)
\(\dfrac{1}{1794}\)>\(\dfrac{1}{1797^2}\)
.....................
\(\dfrac{1}{1794}\)>\(\dfrac{1}{2016^2}\)
\(\dfrac{1}{1794}\)>\(\dfrac{1}{2017^2}\)
\(\Leftrightarrow\)\(\dfrac{1}{1794}\)>\(\dfrac{1}{1795^2}\)+\(\dfrac{1}{1796^2}\)+\(\dfrac{1}{1797^2}\)+. . .+\(\dfrac{1}{2016^2}\)+\(\dfrac{1}{2017^2}\)
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