Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+..........+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{11.12}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...........+\frac{1}{11}-\frac{1}{12}\)
\(=1-\frac{1}{12}\)
\(=\frac{11}{12}\)
1/1.2+1/2.3+1/3.4+1/4.5+...+1/9.10
1-1/2+1/2-1/3+...+1/9-1/10
1-1/10
9/10
đúng 100% đấy
Bài làm
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\frac{1}{1}-\frac{1}{11}\)
\(A=\frac{10}{11}\)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{9\cdot10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}\)
\(B=\frac{9}{10}\)
\(\frac{1}{1}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\)
\(=\frac{1}{1}+\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)
\(=\frac{1}{1}+\left(\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}\right)\)
\(=1+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=1+\frac{2}{5}=\frac{7}{5}\)
1/2=1-1/2
5/6=1-1/6
11/12=1-1/12
109/110=1-1/110
A=10-(1/2+1/6+...+1/110)
B=(1/2+1/6+...+1/110)=[(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/9-1/10)+(1/10+1/11)
B=1+1/11
A=10-(1+1/11)=10+1/11
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
= \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
=\(\frac{1}{1}-\frac{1}{10}\)= \(\frac{9}{10}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{4}{10}=\frac{2}{5}\)
Ủng hộ mk nha !!! ^_^
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+....+\frac{1}{9x10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)