Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(=\dfrac{5}{34}\)
Đặt \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\)
\(A=\dfrac{3}{2}-\dfrac{3}{5}+\dfrac{3}{5}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{14}\)
\(A=\dfrac{3}{2}-\dfrac{3}{14}\)
\(A=\dfrac{21}{14}-\dfrac{3}{14}\)
\(A=\dfrac{18}{14}\)
\(A=\dfrac{9}{7}\)
\(A=1\dfrac{2}{7}\)
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: \(x=\frac{1}{9}\)
\(E=\frac{3^2}{8.11}+\frac{3^2}{11.14}+......+\frac{3^2}{197.200}\)
\(E=3.\left(\frac{3}{8.11}+\frac{3}{11.14}+......+\frac{3}{197.200}\right)\)
\(E=3.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+.....+\frac{1}{197}-\frac{1}{200}\right)\)
\(E=3.\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(E=3.\frac{3}{25}\)
\(E=\frac{9}{25}\)
\(\frac{1}{3}E=\frac{3}{8x11}+\frac{3}{11x14}+..+\frac{3}{197x200}\)
\(\frac{1}{3}E=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+.+\frac{1}{197}-\frac{1}{200}\)
\(\frac{1}{3}E=\frac{1}{8}-\frac{1}{200}\)
\(\frac{1}{3}E=\frac{3}{25}\)
\(E=\frac{9}{25}\)
b)
S2=6/2x5+6/5x8+6/8x11+...+6/29x32
=2.(3/2.5+3/5.8+...+3/29.32)
=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)
=2.(1/2-1/32)
=2.15/32
=15/16
a)
Ta có:
S1=2/3x5+2/5x7+2/7x9+...+2/97x99
=1/3-1/5+1/5-1/7+...+1/97-1/99
=1/3-1/99
=32/99
\(B=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{26\cdot29}\)
\(B=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{29}\)
\(B=\dfrac{1}{2}-\dfrac{1}{29}\)
\(B=\dfrac{27}{58}\)
B= 3/2x5 + 3/5x8+ 3/8x11 + ... + 3/26x29
B= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/26 - 1/29
B= 1/2-1/29
B=27/58
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(=\frac{1}{5}-\frac{1}{2009}\)
\(=\frac{2004}{10045}\)
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 \(A=\frac{2004}{10045}.\)
Đặt A = \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{605.608}\)
\(\Rightarrow3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{605.608}\)
\(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{605}-\frac{1}{608}\)
\(3A=\frac{1}{5}-\frac{1}{608}\)
\(A=\left(\frac{1}{5}-\frac{1}{608}\right).\frac{1}{3}=\frac{201}{3040}\)
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