\(\frac{1}{-2}\cdot \frac{1}{3}+\frac{1}{-3}\cdot \frac{1}{4}+\dots +\frac{1}{-9}\cdot \frac{1}{10}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}\)
\(\Rightarrow-A=\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+...+\dfrac{10-9}{9.10}=\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)
\(=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\Rightarrow A=-\dfrac{2}{5}\)
\(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}\)\(+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(=\frac{1}{4}+\frac{1}{3}\)
\(=\frac{7}{12}\)
1/2.1/3+1/3.1/4+1/4.1/5+1/5.1/6
=1/2.3+1/3.4+1/4.5+1/5.6
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=1/2-1/6
=2/6=1/3
Bài 1:
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+....+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{4033}{2017}\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{4033}{4034}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{4033}{4034}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4033}{4034}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{4033}{4034}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{4033}{4034}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4034}\)
\(\Rightarrow x+1=4034\)
\(\Rightarrow x=4034-1\)
\(\Rightarrow x=4033\)
A = \(\frac{1}{-2}.\frac13+\frac13.\frac{1}{-4}\) + ...+ \(\frac{1}{-9}.\frac{1}{10}\)
A = -(\(\frac12.\) \(\frac13\) + \(\frac13\).\(\frac14\) + ... + \(\frac19\).\(\frac{1}{10}\))
A = -(\(\frac12-\frac13+\frac13-\frac14+\cdots+\frac19-\frac{1}{10}\))
A = - (\(\frac12\) - \(\frac{1}{10}\))
A = - \(\frac25\)