\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2017}\left(1+2+...+2017\right)\)

\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{2017}.\frac{2017\left(2017+1\right)}{2}\)

\(=1+\frac{2.3}{2.2}+\frac{3.4}{3.2}+....+\frac{2017.2018}{2017.2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2018}{2}\)

\(=\frac{2+3+4+...+2018}{2}\)

\(=\frac{\frac{2018\left(2018+1\right)}{2}-1}{2}\)

\(=1018585\)

Suy ra A=1+1.5+2+....+1009=1 013 532.5

Ta có: \(A=\frac{\left(1+\frac{2017}{1}\right)\left(1+\frac{2017}{2}\right)...\left(1+\frac{2017}{1009}\right)}{\left(1+\frac{1009}{1}\right)\left(1+\frac{1009}{2}\right)...\left(1+\frac{1009}{2017}\right)}=\frac{\frac{2017+1}{1}\frac{2017+2}{2}...\frac{2017+1009}{1009}}{\frac{1009+1}{1}\frac{1009+2}{2}...\frac{1009+2017}{2017}}\)

\(\Leftrightarrow A=\frac{\frac{2018.2019...3026}{1.2...1009}}{\frac{1010.1011...3026}{1.2...2017}}=\frac{2018.2019...3026}{1.2...1009}.\frac{1.2...2017}{1010.1011...3026}\)

\(\Leftrightarrow A=\frac{1.2...2017.2018.2019...3026}{1.2...1009.1010.1011...3026}=\frac{1.2.3...3026}{1.2.3...3026}=1.\)

\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{2016}-1\right)\left(\frac{1}{2017}-1\right)\)

\(=\frac{-1}{2}.\frac{-2}{3}...\frac{-2015}{2016}.\frac{-2016}{2017}\)

\(=\frac{1.2...2015.2016}{2.3...2016.2017}\) ( tử số có chẵn số hạng )

\(=\frac{1}{2017}\)

B=1-1/2017

B=2016/2017

(Vì ta đưa bài toán về cách khử đi .VD:1/2+1/4+1/8=1-1/2+1/2-1/4+1/4-1/8

(Tui mới học lớp 5 thôi)

Giải

B=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)+...\left(1-\frac{1}{1+2+3+...+2017}\right)\)

   =\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{\left(1+2017\right).2017:2}\right)\)

   =\(\frac{2}{3}.\frac{5}{6}...\frac{2018.2017:2-1}{2018.2017:2}\)

   =\(\frac{4}{6}.\frac{10}{12}...\frac{\left(2018.2017:2-1\right).2}{2018.2017}\)

   =\(\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{2016.2019}{2017.2018}\)

   =\(\frac{1.2...2016}{3.4....2018}.\frac{4.5...2019}{2.3...2017}\)

   =\(\frac{2}{2017.2018}.\frac{2018.2019}{2.3}\)

   =\(\frac{2019}{2017.3}=\frac{2019}{6051}=\frac{613}{2017}\)

a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)

b, c cùng 1 câu phải k

ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)

\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)

A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)

NHA

HỌC TỐT

\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right)\)

\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{2016.2018}{2017^2}\)

\(=\frac{2.3^2.4^2.5^2...2016^2.2017.2018}{2^2.3^2.4^2.5^2...2017^2}\)

\(=\frac{2018}{2.2017}=\frac{1009}{2017}\)

lơp 6  ko bt

A= E387E4837

B = 883433

C = UỲUWFHQWURY48E3947