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

\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{2019}{2}=1+\frac{3+4+...+2019}{2}=1+\frac{\left(3+2019\right)2017}{2}=2039188\)

thank you bạn

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

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

Tích trên là tích của các thừa số âm và có (2018-2)+1=2017 thừa số nên có kq âm

\(=-\frac{1}{2018}\)

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

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2016}{2017}.\frac{2017}{2018}\)

\(=\frac{1}{2018}\)

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

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2016}{2017}.\frac{2017}{2018}\)

\(=\frac{1.2.3.4....2016.2017}{2.3.4.5....2017.2018}=\frac{1}{2018}\)

phép toán là trừ hay cộng hay nhân ,chia vậy b

ko ghi đề

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

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

=[\(\left(\frac{2}{2.3}\right).\left(\frac{2}{3.4}\right).....\left(\frac{2}{2018.2019}\right)\)]:2

=\(\frac{2^{1008}}{2.3.3.4.....2018.2019}\)

Đoạn này thì ko lm đc nx

nhanh nha mấy bạn mình đang cần rất gấp

\(A=\left(1-\frac{1}{2018}\right)\left(1-\frac{2}{2018}\right)\left(1-\frac{3}{2018}\right)...\left(1-\frac{2020}{2018}\right).\)

   \(=\frac{2017}{2018}\cdot\frac{2016}{2018}\cdot\frac{2015}{2018}\cdot...\cdot\left(1-\frac{2018}{2018}\right)\cdot...\cdot\frac{-2}{2018}\)

   \(=\frac{2017}{2018}\cdot\frac{2016}{2018}\cdot\frac{2015}{2018}\cdot...\cdot0\cdot...\cdot\frac{-2}{2018}\)

   \(=0\)