Cho :
\(M=133.\left(\frac{1}{1.1996}+\frac{1}{2.1997}+\frac{1}{3.1998}+...+\frac{1}{21.2016}\right)\)
\(N=\frac{7}{5}.\left(\frac{1}{1.22}+\frac{1}{2.23}+...+\frac{1}{1995.2016}\right)\)
So sánh M và N
K
Khách
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XO
15 tháng 8 2020
Ta có A = \(133\left(\frac{1}{1.1996}+\frac{1}{2.1997}+...+\frac{1}{17.2002}\right)\)
=> 1995A = \(133\left(\frac{1995}{1.1996}+\frac{1995}{2.1997}+...+\frac{1995}{17.2002}\right)\)
=> 1995A = \(133\left(1-\frac{1}{1996}+\frac{1}{2}-\frac{1}{1997}+...+\frac{1}{17}-\frac{1}{2002}\right)\)
=> 1995A = \(133\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2002}\right)\right]\)
=> A = \(\frac{1}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2002}\right)\right]\)(1)
Lại có B = \(\frac{17}{15}\left(\frac{1}{1.18}+\frac{1}{2.19}+...+\frac{1}{1995.2012}\right)\)
=> 17B = \(\frac{17}{15}\left(\frac{17}{1.18}+\frac{17}{2.19}+...+\frac{17}{1995.2012}\right)\)
=> 17B = \(\frac{17}{15}\left(1-\frac{1}{18}+\frac{1}{2}-\frac{1}{19}+...+\frac{1}{1995}-\frac{1}{2012}\right)\)
=> 17B = \(\frac{17}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{1995}\right)-\left(\frac{1}{18}+\frac{1}{19}+...+\frac{1}{2012}\right)\right]\)
=> 17B = \(\frac{17}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2012}\right)\right]\)
=> B = \(\frac{1}{15}\left[\left(1+\frac{1}{2}+...+\frac{1}{17}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+...+\frac{1}{2012}\right)\right]\)(2)
Từ (1) và (2) => A = B
HT
26 tháng 8 2015
M = \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{2015^2}\right)\)
M = \(\left(-\frac{1.3}{2.2}\right)\left(-\frac{2.4}{3.3}\right)\left(-\frac{3.5}{4.4}\right)....\left(-\frac{2014.2016}{2015.2015}\right)\)
M = \(\frac{\left(1.2.3....2014\right)\left(3.4.5...2016\right)}{\left(2.3.4.....2015\right)\left(2.3.4....2015\right)}\)
M = \(\frac{2016}{2015.2}\)
M = \(\frac{1008}{2015}\)
N = \(\frac{1}{2}\)=\(\frac{1008}{2016}\)
Vì \(\frac{1008}{2015}>\frac{1008}{2016}\)
=> M > N
11 tháng 8 2016
\(\sqrt{1.1998}< \frac{1+1998}{2}\)
\(S>\frac{2}{1999}+\frac{2}{1999}+...+\frac{2}{1999}=2.\frac{1998}{1999}\)
PL
12 tháng 3 2020
\(2\frac{1998}{1999}\)là hỗn số hay \(2.\frac{1998}{1999}\)hả bạn?
WH
1 tháng 5 2018
Ta có
\(A=\frac{\left(3\frac{2}{5}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\) \(B=\frac{1,2:\left(1\frac{1}{5}-1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
\(\Leftrightarrow A=\frac{\left(\frac{17}{5}+\frac{1}{5}\right):\frac{5}{2}}{\left(\frac{38}{7}-\frac{9}{4}\right):\frac{276}{56}}\) \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(\frac{6}{5}-\frac{5}{4}\right)}{\frac{8}{25}+\frac{2}{25}}\)
\(\Leftrightarrow A=\frac{\frac{18}{5}:\frac{5}{2}}{\frac{89}{28}:\frac{276}{56}}\) \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(-\frac{1}{20}\right)}{\frac{2}{5}}\)
\(\Leftrightarrow A=\frac{\frac{36}{25}}{\frac{89}{138}}\) \(\Leftrightarrow B=\frac{\frac{5}{4}}{\frac{2}{5}}\)
\(\Leftrightarrow A=\frac{4968}{2225}\) \(\Leftrightarrow B=\frac{25}{8}\)
\(\Leftrightarrow A=\frac{39744}{17800}\) \(\Leftrightarrow B=\frac{55625}{17800}\)
Ta có: 39744<55625
\(\Rightarrow A< B\)
Vậy A<B
10 tháng 7 2018
M=-(\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{1-100^2}{100^2}\))
=-(\(\frac{1.3}{2.2}.\frac{2.4}{3.3}\frac{3.5}{4.4}...\frac{99.100}{100.100}\))
=-(\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...100}{2.3.4..100}\))
=-(\(\frac{1}{100}.\frac{1}{2}\))
=\(\frac{-1}{200}\)
Ta có :
\(M=133.\left(\frac{1}{1.1996}+\frac{1}{2.1997}+..........+\frac{1}{21.2016}\right)\)
\(\Rightarrow M.15=133.15.\left(\frac{1}{1.1996}+\frac{1}{2.1997}+.......+\frac{1}{21.2016}\right)\)
\(\Rightarrow M.15=\frac{1995}{1.1996}+\frac{1995}{2.1997}+........+\frac{1995}{21.2016}\)
\(\Rightarrow M.15=1-\frac{1}{1996}+\frac{1}{2}-\frac{1}{1997}+...........+\frac{1}{21}-\frac{1}{2016}\)
\(\Rightarrow M.15=\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{21}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+.....+\frac{1}{2016}\right)\)
Ta có:
\(N.15=\frac{7}{5}.15\left(\frac{1}{1.22}+\frac{1}{2.23}+..........+\frac{1}{1995.2016}\right)\)
\(\Rightarrow N.15=\frac{21}{1.22}+\frac{21}{2.23}+..........+\frac{21}{1995.2016}\)
\(\Rightarrow N.15=1-\frac{1}{22}+\frac{1}{2}-\frac{1}{23}+.............+\frac{1}{1995}-\frac{1}{2016}\)
\(\Rightarrow N.15=\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{1995}\right)-\left(\frac{1}{22}+\frac{1}{23}+.......+\frac{1}{2016}\right)\)
\(\Rightarrow N.15=\left(1+\frac{1}{2}+.....+\frac{1}{21}\right)+\left(\frac{1}{22}+\frac{1}{23}+....+\frac{1}{1995}-\frac{1}{22}-...-\frac{1}{2016}\right)\)
\(\Rightarrow N.15=\left(1+\frac{1}{2}+....\frac{1}{21}\right)-\left(\frac{1}{1996}+\frac{1}{1997}+....\frac{1}{2016}\right)\)
\(\Rightarrow N.15=M.15\Rightarrow M=N\)
soyeon_Tiểubàng giải
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