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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
Võ Đông Anh Tuấn
Silver bullet
Hoàng Lê Bảo Ngọc
Trần Việt Linh
Lê Nguyên Hạo
mấy bn giúp mk vs
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
cho công thức tổng quát nè (do tui tự nghĩ ra đó :))
\(\left(a+b\right)\left(\frac{1}{?}\right)=\frac{a+b}{?}\) dựa vô đây nhân (1+17) và (1+19) và từng cái ngoặc kia là đc
\(A=\frac{\frac{1+17}{1}\cdot\frac{2+17}{2}\cdot\frac{3+17}{3}\cdot...\cdot\frac{19+17}{19}}{\frac{1+19}{1}\cdot\frac{2+19}{2}\cdot\frac{3+19}{3}\cdot...\cdot\frac{17+19}{17}}=\frac{18\cdot19\cdot20\cdot...\cdot36}{1\cdot2\cdot3\cdot...\cdot19}:\frac{20\cdot21\cdot22\cdot...\cdot36}{1\cdot2\cdot3\cdot...\cdot17}\)
\(=\frac{18\cdot19}{18\cdot19}=1\)
Ta có:
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)
\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)
\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)
\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)
\(A=\frac{1.2018}{2017.2}\)
\(A=\frac{1009}{2017}\)
Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)
\(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)
Vậy A>B
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\left(\frac{-15}{16}\right)\cdot\cdot\cdot\left(\frac{-4052168}{4052169}\right)\left(\frac{-4056195}{4056196}\right)\)
\(A=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot....\cdot\frac{-2012\cdot2014}{2013\cdot2013}\cdot\frac{-2013\cdot2015}{2014\cdot2014}\)
\(A=\frac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2012\right)\cdot\left(-2013\right)}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\cdot\frac{3\cdot4\cdot5\cdot....\cdot2014\cdot2015}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\)
\(A=\frac{-1}{2014}\cdot\frac{2015}{2}=\frac{-2015}{4028}\)
Ta thấy \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)
Ta có : \(\frac{1}{n^2}-1=\frac{1-n^2}{n^2}=\frac{\left(1-n\right)\left(1+1\right)}{n^2}\)
Áp dụng :
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}.....\frac{-2013.2015}{2014.2014}\)
\(=\frac{-\left(1.2.3...2013\right)\left(3.4.5....2015\right)}{\left(2.3.4.....2014\right)\left(2.3.4......2014\right)}=\frac{-2015}{2014.2}=\frac{-2015}{4028}\)
Sr còn thiếu
\(A=-\frac{2015}{4028}< \frac{-2014}{4028}=-\frac{1}{2}\)
Vậy \(A< B\)
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