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\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+\frac{2}{2018}+\frac{3}{2017}+...+\frac{2018}{2}+\frac{2019}{1}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+1+\frac{2}{2018}+1+\frac{3}{2017}+1+...+\frac{2018}{2}+1+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{2020}{2019}+\frac{2020}{2018}+\frac{2020}{2017}+...+\frac{2020}{2}+\frac{2020}{2020}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}\right)}\)
\(\frac{A}{B}=\frac{1}{2020}\)
Ta có 200920= 20092x10=(20092)10= 403608110
Vì 4036081<20092009
Nên 403608110<2009200910
Vậy...
Rồi đó nha
~ủng hộ dùm~
ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1
=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2
=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019
=> B= 2019 *(1/2+1/3+...+1/2019)
=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)
=> A/B= 1/2019
\(\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}+2018}{\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(=\frac{1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)+2018}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}+\frac{2018}{1}}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(=\frac{2018.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
= 2018
\(\text{a,}\frac{2}{13}.\frac{-5}{3}+\frac{11}{13}.\frac{-5}{3}=-\frac{5}{3}\left(\frac{2}{13}+\frac{11}{13}\right)\)
\(=\frac{-5}{3}.\frac{13}{13}\)
\(=-\frac{5}{3}\)
\(\text{b,}\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^3.27+\left(\frac{-2017}{2018}\right)^0=\frac{1}{9}-\frac{1}{27}.27+1\)
\(=\frac{1}{9}-1+1\)
\(=\frac{1}{9}\)
\(\text{c,}1,2-\sqrt{\frac{1}{4}}:1\frac{1}{20}+\left|\frac{3}{4}-1,25\right|-\left(\frac{-3}{2}\right)^2=\frac{6}{5}-\frac{1}{2}:\frac{21}{20}+\left|\frac{3}{4}-\frac{5}{4}\right|-\frac{9}{4}\)
\(=\frac{6}{5}-\frac{10}{21}+\frac{1}{2}-\frac{9}{4}\)
\(=\frac{-431}{420}\)
\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).....\left(1-\frac{1}{1+2+3+.....+2018}\right)\)
\(=\left(1-\frac{1}{\frac{2.3}{2}}\right).\left(1-\frac{1}{\frac{3.4}{2}}\right).......\left(1-\frac{1}{\frac{2018.2019}{2}}\right)\)
\(=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).......\left(1-\frac{2}{2018.2019}\right)\)
\(=\left(1-\frac{1}{3}\right).\left(1-\frac{5}{6}\right).......\left(1-\frac{1}{2037171}\right)\)
\(=\frac{2}{3}.\frac{5}{6}......\frac{2037170}{2037171}\)
\(=\frac{4}{6}.\frac{10}{12}.......\frac{4074340}{4074342}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}......\frac{2017.2020}{2018.2019}\)
\(=\frac{1.2......2017}{2.3.....2018}.\frac{4.5......2020}{3.4......2019}=\frac{1}{2018}.\frac{2020}{3}=\frac{1010}{3027}\)
\(6.\left(-\frac{1}{3}\right)^2-\frac{5}{4}:0,5+3\frac{1}{2}\)
\(=6.\frac{1}{9}-\frac{5}{4}.2+\frac{7}{2}\)
\(=\frac{2}{3}-\frac{5}{2}+\frac{7}{2}\)
\(=-\frac{11}{6}+\frac{7}{2}\)
\(=\frac{5}{3}\)
\(\frac{2017}{2018}.\frac{15}{17}-\frac{32}{17}.\frac{2017}{2018}=\frac{2017}{2018}.\left(\frac{15}{17}-\frac{32}{17}\right)\)
\(=\frac{2017}{2108}.\left(-1\right)=-\frac{2017}{2018}\)
\(\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right).....\left(-\frac{2017}{2018}\right)=\frac{-1.\left(-2\right).....\left(-2017\right)}{2.3.....2018}=-\frac{1}{2018}\)
đặt \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{2018}-1\right)\)
\(-A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2018}\right)\)
\(-A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2017}{2018}\)
\(-A=\frac{1}{2018}\)
\(A=-\frac{1}{2018}\)