Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{605.608}\)
= \(\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{605}-\frac{1}{608}\right)\)
= \(\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{608}\right)\)
= \(\frac{1}{3}.\frac{75}{608}\)
= \(\frac{25}{608}\)
Đặt \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\)
\(A=\dfrac{3}{2}-\dfrac{3}{5}+\dfrac{3}{5}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{14}\)
\(A=\dfrac{3}{2}-\dfrac{3}{14}\)
\(A=\dfrac{21}{14}-\dfrac{3}{14}\)
\(A=\dfrac{18}{14}\)
\(A=\dfrac{9}{7}\)
\(A=1\dfrac{2}{7}\)
\(3\times\left(\frac{1}{5\times8}+\frac{1}{8\times11}+....+\frac{1}{97\times100}+x\right)=\frac{319}{100}\)
\(\Rightarrow\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{97\times100}\right)+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{19}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow3\times x=\frac{319}{100}-\frac{19}{100}\)
\(\Rightarrow3\times x=3\)
\(\Rightarrow x=3:3\)
\(\Rightarrow x=1\)
Vậy x = 1
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{27}{480}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{27}{480}.\frac{1}{3}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{3}{160}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{160}\)
\(\Rightarrow\frac{1}{x+3}=\frac{31}{160}\)
\(\Rightarrow160=31x+93\)
\(\Rightarrow31x=67\)
\(\Rightarrow x=\frac{67}{31}\)
A = \(\dfrac{1}{2}\) x 5 + \(\dfrac{1}{5}\) x 8 + \(\dfrac{1}{8}\) x 11 + \(\dfrac{1}{14}\) x 17
A = \(\dfrac{5}{2}\) + \(\dfrac{8}{5}\) + \(\dfrac{11}{8}\) + \(\dfrac{17}{14}\)
A = \(\dfrac{700}{280}\) + \(\dfrac{448}{280}\) + \(\dfrac{385}{280}\) + \(\dfrac{340}{280}\)
\(\Rightarrow\) A = \(\dfrac{1873}{280}\)
A \(=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}\)
A \(=\)\(\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}\right)\)
A \(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\right)\)
A \(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{17}\right)\)
A \(=\dfrac{1}{3}.\dfrac{15}{34}\)
A \(=\dfrac{5}{34}\)
Ta có: \(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\frac{3}{7}=\frac{1}{21}\)
\(\Leftrightarrow x=\frac{1}{21}:\frac{3}{7}=\frac{1}{21}\cdot\frac{7}{3}=\frac{7}{63}=\frac{1}{9}\)
Vậy: \(x=\frac{1}{9}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(=\frac{1}{5}-\frac{1}{2009}\)
\(=\frac{2004}{10045}\)
Đề: Tính
\(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)
Vậy \(A=\frac{2004}{10045}.\)
top scorer cop tại:tính nhanh:2/2*5+2/5*8+2/8*11+2/11*14+2/14*17? | Yahoo Hỏi & Đáp
có cách làm tại:Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
Đặt A = \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{605.608}\)
\(\Rightarrow3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{605.608}\)
\(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{605}-\frac{1}{608}\)
\(3A=\frac{1}{5}-\frac{1}{608}\)
\(A=\left(\frac{1}{5}-\frac{1}{608}\right).\frac{1}{3}=\frac{201}{3040}\)