B=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......................+\frac{4}{999=1000}\)
Những câu hỏi liên quan
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}............+\frac{1}{999+1000}+1\)1
=1-1/2+1/2-1/3+...+1/999-1/1000+1
=1-1/100+1
=199/100
Đúng 0
Bình luận (0)
\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{999^2}{999\cdot1000}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{999^2}{999.1000}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.....\frac{999.999}{999.1000}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{999}{1000}\)
\(=\frac{1}{1000}\)
Đúng 0
Bình luận (0)
Tính tổng :a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
Tính tổng :
a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}
b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}
c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}
d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}
e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
Đúng 0
Bình luận (0)
Tính nhanh:
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{1000\cdot1001}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1000\cdot1001}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1000}-\frac{1}{1001}\)
\(=1-\frac{1}{1001}\)
\(=\frac{1000}{1001}\)
Đúng 0
Bình luận (0)
1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+.............+\(\frac{1}{1000}\)-\(\frac{1}{1001}\)=1-\(\frac{1}{1001}\)=\(\frac{1000}{1001}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Bài 1:
a) \(\frac{1}{1}\cdot2+\frac{1}{2}\cdot3+\frac{1}{3}\cdot4+...+\frac{1}{n}\cdot\left(n+1\right)\)
b) \(\frac{1}{1}\cdot2\cdot3+\frac{1}{2}\cdot3\cdot4+\frac{1}{3}\cdot4\cdot5+...+\frac{1}{a}\cdot\left(a+1\right)\cdot\left(a+2\right)\)
\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+.....+\frac{2016}{501}}{\frac{-1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-.....-\frac{1}{999\cdot1000}}\)
\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+...+\frac{2016}{501}}{-\frac{1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-...-\frac{1}{999\cdot1000}}\)
\(B=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{999\cdot1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{500}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}\)
\(B=\frac{2016}{-1}=-2016\)
Đúng 0
Bình luận (0)
1.a) frac{1}{1cdot2}+frac{1}{2cdot3}+frac{1}{3cdot4}+....+frac{1}{2017cdot1018}b) frac{2}{1cdot2}+frac{2}{2cdot3}+frac{2}{3cdot4}+....+frac{2}{2017cdot2018}+frac{2}{2018cdot2019}c) frac{4}{1cdot2}+frac{4}{2cdot3}+frac{4}{3cdot4}+......+frac{4}{1999cdot2000}********Lưu ý :1) Các dấu chấm ở phân số là dấu nhân ạ..!!!2) Trình bày rõ ràng giúp mk với ạ..!!!
Đọc tiếp
1.
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2017\cdot1018}\)
b) \(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+....+\frac{2}{2017\cdot2018}+\frac{2}{2018\cdot2019}\)
c) \(\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+......+\frac{4}{1999\cdot2000}\)
********Lưu ý :
1) Các dấu chấm ở phân số là dấu nhân ạ..!!!
2) Trình bày rõ ràng giúp mk với ạ..!!!
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)
b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(1-\frac{1}{2019}\right)\)
\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2.\frac{2018}{2019}\)
\(=\frac{4036}{2019}\)
Phần c tương tự nha
Đúng 0
Bình luận (0)
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + .......+ \(\frac{1}{2017.2018}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + .......+ \(\frac{1}{2017}\) - \(\frac{1}{2018}\)
= 1 - \(\frac{1}{2018}\) = \(\frac{2017}{2018}\)
câu a) mik sửa đề một tí ko biết có đúng ko
câu b , c tương tự nhưng cần lấy tử ra chung
Đúng 0
Bình luận (0)
a)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b)nhóm 2 ra ngoài rồi làm như câu a
c)nhóm 4 ra rồi làm như câu a
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Chứng minh rằng
a, B = \(\frac{1\cdot2-1}{2!}+\frac{2\cdot3-1}{3!}+\frac{3\cdot4-1}{4!}+....+\frac{99\cdot100-1}{100!}< 2\)
c, C = \(\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{19}{9^2\cdot10^2}< 1\)
\(C=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+....+\frac{99.100-1}{100!}\)
\(\Rightarrow C=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(\Rightarrow C=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(\Rightarrow C=\left(2+\frac{3.4}{4!}+\frac{4.5}{5!}+....+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{10!}\right)\)
\(\Rightarrow C=\left(2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(\Rightarrow C=2-\frac{1}{99!}-\frac{1}{100!}< 2\Rightarrow C< 2\)
\(b,C=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+....+\frac{19}{9^2.10^2}\)
\(\Rightarrow C=\frac{3}{\left(1.2\right)\left(1.2\right)}+\frac{5}{\left(2.3\right)\left(2.3\right)}+...+\frac{19}{\left(9.10\right)\left(9.10\right)}\)
\(\Rightarrow C=\frac{3}{1.2}.\frac{1}{1.2}+\frac{5}{2.3}.\frac{1}{2.3}+....+\frac{19}{9.10}.\frac{1}{9.10}\)
\(\Rightarrow C=\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{9}+\frac{1}{10}\right)\left(\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{90}\)
\(\Rightarrow C=1-\frac{1}{90}< 1\Rightarrow C< 1\)
Đúng 0
Bình luận (0)