Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
a)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+\frac{5-3}{3\cdot4\cdot4}+....+\frac{100-98}{98\cdot99\cdot100}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{35.37}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{35}-\frac{1}{37}\)
\(=\frac{1}{3}-\frac{1}{37}=\frac{34}{111}\)
c) \(\frac{7}{7.9}+\frac{7}{9.11}+\frac{7}{11.13}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{101}\right)=\frac{7}{2}\cdot\frac{94}{707}=\frac{47}{101}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{1}-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
\(\:\frac{-1}{1.2}+\frac{-1}{2.3}+\frac{-1}{3.4}+\frac{-1}{4.5}\)
\(=-1\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)\)
=\(-1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
=\(-1\left(1-\frac{1}{5}\right)\)
=\(-1\times\frac{4}{5}\)
=\(\frac{-4}{5}\)
Thay n = a nha / lúc trước có giải r nên ko giải lại rắc rối
\(N=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{a\left(a+1\right)}\)
\(\Rightarrow N=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{a}-\frac{1}{a-1}\)
\(\Rightarrow N=1-\frac{1}{a-1}\)
\(\Rightarrow N=\frac{a-1-1}{a-1}\)
\(\Rightarrow N=\frac{a-2}{a-1}\)
Đặt A=(1-2/2*3)*(1-2/3*4)*(1-2*/4*5)*...*(1-2/99*100)
A=\(\frac{1\cdot4}{2\cdot3}\)*\(\frac{2\cdot5}{3\cdot4}\)*\(\frac{3\cdot6}{4\cdot5}\)*...*\(\frac{98\cdot101}{99\cdot100}\)
A=\(\frac{1\cdot101}{3\cdot99}\)
A=\(\frac{101}{297}\)
\(-\frac{1}{1.2}+-\frac{1}{2.3}+-\frac{1}{3.4}+-\frac{1}{4.5}\)
\(=-1\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)\)
\(=-1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
\(=-1\left(1-\frac{1}{5}\right)\)
\(=-1.\frac{4}{5}=-\frac{4}{5}\)
\(\frac{-1}{1.2}+\frac{-1}{2.3}+\frac{-1}{3.4}+\frac{-1}{4.5}\)
\(=-1\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)\)
\(=-1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
\(=-1\left(1-\frac{1}{5}\right)\)
\(=-1.\frac{4}{5}=-\frac{4}{5}\)