\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2017.2019}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{2017.2019+1}{2017.2019}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2018^2}{2017.2019}\)
\(=\frac{2}{1}.\frac{2018}{2019}=\frac{4036}{2019}\)
cách bạn Chi hay nè
Thực hiện phép tinh sau:
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\)
(1+\(\frac{1}{1.3}\)).(1+\(\frac{1}{2.4}\))+(1+\(\frac{1}{3.5}\)).......(1+\(\frac{1}{2017.2019}\))
Thực hiện phép tính:
B =\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
Thực hiện phép tính: \(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)
Thực hiện phép tính :
A = \(\frac{1-3}{1.3}+\frac{2-4}{2.4}+\frac{3-5}{3.5}+..........+\frac{2012-2014}{2012.2014}-\frac{2013+2014}{2013.2014}\)
giúp mình nha
S=\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)
bai 1: s=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
bai 2: s=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\text{+}...\text{+}\frac{1}{2017.2019}\)
Bài 1: Thực hiện phép tính 1 cách hợp lý :
\(a\)) \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2009.2011}\right)\)
\(b\)) \(\frac{16}{9}-\frac{1}{36}-\frac{1}{28}-\frac{1}{21}-\frac{1}{15}-\frac{1}{10}-\frac{1}{6}-\frac{1}{3}-1\)
\(c\)) \(\frac{0,4-\frac{2}{9}+\frac{2}{11}-\frac{2}{37}}{1,4-\frac{7}{9}-\frac{7}{11}-\frac{7}{37}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,785+0,7}\)
Chứng minh rằng: \(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}< \frac{3}{4}\)
