NH
1
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
AM
14 tháng 6 2015
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
VM
2
LS
10 tháng 7 2018
Đặt A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+10}\)
\(A=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{10.11}{2}}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(A=2\cdot\frac{9}{22}=\frac{9}{11}\)
Vậy A = \(\frac{9}{11}\)
NT
2
18 tháng 7 2015
\(B=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{8.9.10.11}+\frac{1}{9.10.11.12}\)
\(B=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{8.9.10.11}+\frac{3}{9.10.11.12}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{9.10.11}-\frac{1}{10.11.12}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{10.11.12}\right)\)
\(B=\frac{1}{3}.\frac{73}{440}=\frac{43}{1320}\)
25 tháng 6 2018
\(5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2\frac{1}{2}\right):\frac{7}{4}\)
=\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{9}{2}-\frac{5}{2}\right):\frac{7}{4}\)
=\(\frac{118}{30}-\left(\frac{63}{6}-\frac{5}{2}\right)x\frac{4}{7}\)
=\(\frac{118}{30}-\left(\frac{63}{6}-\frac{15}{6}\right)x\frac{4}{7}\)
=\(\frac{118}{30}-8x\frac{4}{7}\)
=\(\frac{118}{30}-\frac{32}{7}\)
sau bn tự tính nha
DN
18 tháng 8 2015
1/1.2+1/2.3+1/3.4+...+1/9.10=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10=1-1/10=9/10
5 tháng 8 2019
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
=1-\(\frac{1}{10}\)
=\(\frac{9}{10}\)
5 tháng 8 2019
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)
=\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}\)
=\(2\cdot\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
=\(2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=2\cdot\left(\frac{1}{2}-\frac{1}{11}\right)=\frac{9}{11}\)