Ta có:

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

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

Đởn giản hết sẽ còn là:

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

có ai biết câu a, ko vậy

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

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

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

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

lơp 6  ko bt

mk mới học lớp 6 thôi

tk mk nha mk đang âm điểm nè hu hu

1

tick mình nha thank

\(\frac{\left(1+\frac{2017}{1}\right)\left(1+\frac{2017}{2}\right)....\left(1+\frac{2017}{1009}\right)}{\left(1+\frac{1009}{1}\right)\left(1+\frac{1009}{2}\right)....\left(1+\frac{1009}{2017}\right)}=\frac{1.1.1.....1}{1.1.1....1}=1\)

- Đề sai rồi : )
- Xem lại đề nha bạn #Thành

A=\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2017.2019}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)..........\left(\frac{2017.2019+1}{2017.2019}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.............\frac{4072324}{2017.2019}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...................\frac{2018^2}{2017.2019}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{\left(2.3.4..........2018\right).\left(2.3.4............2018\right)}{\left(1.2.3............2017\right).\left(3.4.5..........2019\right)}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2018.2}{1.2019}\right)=\frac{2018.2}{2.2019}=\frac{2018}{2019}\)

Vậy \(A=\frac{2018}{2019}\)

Chúc bn học tốt

\(A:\frac{1}{2}=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2017.2019+1}{2017.2019}\)

\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{2018^2}{2017.2019}\)

\(=\frac{2.2.3.3.4.4.....2018.2018}{1.3.2.4.3.5....2017.2019}\)

\(=\frac{2.3.4.....2018}{1.2.3.4.....2017}.\frac{2.3.4....2018}{3.4.5.....2019}\)

\(=2018.\frac{2}{2019}\)

\(=\frac{4036}{2019}\)

\(\Rightarrow A=\frac{4036}{2019}.\frac{1}{2}\)

\(A=\frac{2018}{2019}\)

C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!

Câu hỏi của Kurosaki Akatsu - Toán lớp 8 - Học toán với OnlineMath

Bài này tôi đang ôn trong quyển học sinh giỏi nè

Với mọi n thuộc N^* ta có :

\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)

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

\(\Rightarrow N=\frac{\left(2^2+2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)...\left(2008^2+2008+\frac{1}{2}\right)\left(2008^2-2008+\frac{1}{2}\right)}{\left(1^2+1+\frac{1}{2}\right)\left(1^2-1+\frac{1}{2}\right)...\left(2007^2+2007+\frac{1}{2}\right)\left(2007^2-2007+\frac{1}{2}\right)}\)

\(=\frac{\left(2.3+\frac{1}{2}\right)\left(1.2+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)...\left(2008.2009+\frac{1}{2}\right)}{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)...\left(2007.2008+\frac{1}{2}\right)}\)

\(=\frac{2008.2009+\frac{1}{2}}{\frac{1}{2}}=8068145\)