Thực hiện phép tính:\(\left(1-\frac{1}{2018}\right).\left(1-\frac{2}{2018}\right).\left(1-\frac{3}{2018}\right)...\left(1-\frac{2020}{2018}\right)\)
Thực hiện phép tính : \(\left(1\frac{1}{1+2}\right).\left(1\frac{1}{1+2+3}\right)...\left(1\frac{1}{1+2+3+...+2018}\right)\)
Thực hiện phép tính:
\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).....\left(1-\frac{1}{1+...+2018}\right)\)
Thực hiện phép tính
\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2018}\right)\)
\(A=\)\(\left(1-\frac{1}{2018}\right)\)\(\left(1-\frac{2}{2018}\right)\)\(\left(1-\frac{3}{2018}\right)\)\(...\)\(\left(1-\frac{2020}{2018}\right)\)
\(A=\left(1-\frac{1}{2018}\right)\left(1-\frac{2}{2018}\right)\left(1-\frac{3}{2018}\right)...\left(1-\frac{2020}{2018}\right).\)
\(=\frac{2017}{2018}\cdot\frac{2016}{2018}\cdot\frac{2015}{2018}\cdot...\cdot\left(1-\frac{2018}{2018}\right)\cdot...\cdot\frac{-2}{2018}\)
\(=\frac{2017}{2018}\cdot\frac{2016}{2018}\cdot\frac{2015}{2018}\cdot...\cdot0\cdot...\cdot\frac{-2}{2018}\)
\(=0\)
Tính:\(\left(2018-\frac{1}{3}-\frac{2}{4}-\frac{3}{5}-...-\frac{2018}{2020}\right):\left(\frac{1}{15}+\frac{1}{20}+\frac{1}{25}+...+\frac{1}{10100}\right)\)
nhanh nha mấy bạn mình đang cần rất gấp
thực hiện phép tính:
\(\left(1\frac{1}{1+2}\right).\left(1\frac{1}{1+2+3}\right).....\left(1\frac{1}{1+2+3+...+2018}\right)\)
Tính
\(\left(2018-\frac{1}{3}-\frac{2}{4}-\frac{3}{5}-\frac{4}{6}-...-\frac{2018}{2020}\right):\left(\frac{1}{15}+\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)\)
Hello Triệu Mẫn điên .Tui là Nguyên 6n1^^
Tui đang suy nghĩ
Tui biết làm nhưng không nói
chỉ nói kết quả bằng 10
Thực hiện phép tính :
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
â , tính M = \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right).......\left(1+\frac{1}{2017}\right)\left(1+\frac{1}{2018}\right)\)
b , Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}\)
c , B = \(\frac{1}{1010}+\frac{1}{1011}+.....+\frac{1}{2017}+\frac{1}{2018}.tinh\left(\frac{A}{B}\right)^{2018}\)
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