NL
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
KN
31 tháng 7 2020
\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\frac{8}{33}=\frac{10}{11}.\frac{33}{8}\)
\(=\frac{15}{4}\)
31 tháng 7 2020
Trả lời:
\(\frac{10}{11}\div\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{10}{11}\div\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{10}{11}\div\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{10}{11}\div\frac{8}{33}\)
\(=\frac{10}{11}\times\frac{33}{8}\)
\(=\frac{15}{4}\)
T
1 tháng 8 2020
\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\left(\frac{11}{33}-\frac{3}{33}\right)\)
\(=\frac{10}{11}:\frac{8}{33}\)
\(=\frac{15}{4}\)
Học tốt
1 tháng 8 2020
\(\frac{10}{11}:\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{10}{11}:\frac{8}{33}\)
\(=\frac{10}{11}.\frac{33}{8}\)
\(=\frac{15}{4}\)
30 tháng 7 2020
\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
30 tháng 7 2020
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\\ \)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{4}{8}-\frac{1}{8}\\ =\frac{3}{8}\)
Chúc bn học thiệt giỏi nhé!
11 tháng 6 2016
Mình sửa lại đề xíu.
a) \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}=\frac{3}{4}+\frac{1}{4}+\frac{18}{21}+\frac{3}{21}+\frac{19}{32}+\frac{13}{32}=1+1+1=3\)
b) \(4\frac{2}{5}+5\frac{6}{9}+2\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}=4+\frac{2}{5}+\frac{3}{5}+5+\frac{2}{3}+\frac{1}{3}+2+\frac{3}{4}+\frac{1}{4}\)
\(=4+1+5+1+2+1=14.\)
c) \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{41\cdot43}=\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+...+\frac{43-41}{41\cdot43}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{43}=\frac{1}{3}-\frac{1}{43}=\frac{43-3}{3\cdot43}=\frac{40}{129}.\)
5 tháng 2 2017
a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)= \(\frac{3\cdot4\cdot7}{3\cdot4\cdot8\cdot9}\)= \(\frac{7}{72}\)
b. \(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)= \(\frac{4\cdot5\cdot2\cdot3}{3\cdot4\cdot5\cdot2\cdot8}\)= \(\frac{1}{8}\)
c.\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)= \(\frac{5\cdot6\cdot7}{2\cdot6\cdot2\cdot7\cdot3\cdot5}\)= \(\frac{1}{12}\)
5 tháng 2 2017
a, \(\frac{3.4.7}{12.8.9}\)= \(\frac{3.4.7}{3.4.8.9}\)= \(\frac{7}{72}\)
b, \(\frac{4.5.6}{12.10.8}\)= \(\frac{4.5.6}{3.4.2.5.8}\)= \(\frac{1}{8}\)
c, \(\frac{5.6.7}{12.14.15}\)= \(\frac{5.6.7}{2.6.2.7.3.5}\)= \(\frac{1}{12}\)
25 tháng 4 2016
\(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{99.101.103}\)
=\(\frac{1}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{99.101.103}\right)\)
=\(\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{99.101}-\frac{1}{101.103}\right)\)
=\(\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{101.103}\right)\)
=\(\frac{1}{4}.\frac{10406}{31209}\)
=\(\frac{5230}{62418}\)
DP
24 tháng 6 2017
\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{19.21}\right).x=\frac{9}{7}\)
\(\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{21-19}{19.21}\right).x=\frac{9}{7}\)
\(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)
\(\left[\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\right].x=\frac{9}{7}\)
\(\left[\frac{1}{2}.\frac{2}{7}\right].x=\frac{9}{7}\)
\(\frac{1}{7}.x=\frac{9}{7}\)
\(\Rightarrow x=\frac{9}{7}\div\frac{1}{7}=9\)
24 tháng 6 2017
\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)
\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)x=\frac{9}{7}\)
\(\Leftrightarrow\left(\frac{1}{3}-\frac{1}{21}\right)x=\frac{9}{7}\)
\(\Leftrightarrow\frac{2}{7}x=\frac{9}{7}\)
\(\Leftrightarrow x=\frac{9}{2}\)
4 tháng 5 2018
bài 1
Ta có : 2016/2017<1
2017/2018<1
Nên 2016/2017=2017/2018
4 tháng 5 2018
Bài 1 :
a) Ta có : \(\frac{2016}{2017}=1-\frac{1}{2017}\)
\(\frac{2017}{2018}=1-\frac{1}{2018}\)
Vì \(-\frac{1}{2017}< -\frac{1}{2018}\)nên \(\frac{2016}{2017}< \frac{2017}{2018}\)
b) Ta có : \(\frac{2018}{2017}=1+\frac{1}{2017}\)
\(\frac{2017}{2016}=1+\frac{1}{2016}\)
Vì \(\frac{1}{2017}< \frac{1}{2016}\) nên \(\frac{2018}{2017}< \frac{2017}{2016}\)
Câu 2 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{97\cdot99}\)
=)\(\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+......+\frac{99-97}{97\cdot99}\)
=)\(\frac{5}{3\cdot5}-\frac{3}{3\cdot5}+\frac{5}{5\cdot7}\cdot\frac{7}{5\cdot7}+.......+\frac{99}{97\cdot99}-\frac{97}{97\cdot99}\)
=)\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{97}-\frac{1}{99}\)
=)\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
CHÚC BẠN HỌC TỐT NHÉ
32/99 nhé