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a,
\(\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}........\dfrac{-99}{100}.\dfrac{-120}{121}\)
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.........\dfrac{9.11}{10^2}.\dfrac{10.12}{11^2}\)
\(=\dfrac{1.2.3.4.....10.3.4.5.6......11.12}{2^2.3^2........11^2}\)
\(=\dfrac{1.2.11.12}{2^2.11^2}=\dfrac{12}{22}\)
\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\\ \Rightarrow S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(M=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow S=2^{2010}-M\)
* Tính M
\(M=2^{2009}+2^{2008}+...+2+1\\ \Rightarrow2^0+2^1+...+2^{2008}+2^{2009}\\ \Rightarrow2S=2^1+2^2+...+2^{2009}+2^{2010}\\ \Rightarrow2S-S=\left(2^1+2^2+...+2^{2009}+2^{2010}\right)-\left(2^0+2^1+...+2^{2008}+2^{2009}\right)\\ \Rightarrow S=2^{2010}-2^0=2^{2010}-1\)Thay M vào S, ta được :
\(S=2^{2010}-\left(2^{2010}-1\right)\\ \Rightarrow S=2^{2010}-2^{2010}+1\\ \Rightarrow S=1\)
\( S =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1} {2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right) \)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\) \(\Rightarrow S=P\)\)
\(B=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(B=1+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{1}{2018}+1\right)\)
\(B=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(B=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
ta có \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
