tìm x:\(\frac{32}{8.11}+\frac{32}{11.14}+\frac{32}{14.17}+...+\frac{32}{197.200}-x=\frac{1}{2}\)
[\(\frac{2000}{2000.2006}+\frac{2000}{2006.2012}+\frac{2000}{2012.2018}+.....+\frac{2000}{2492.2498}\)]x\(\frac{^{3^2}}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+.....+\frac{3^2}{197.200}\)
\(\left[\frac{2000}{2000.2006}+\frac{2000}{2006.2012}+...+\frac{2000}{2492.2498}\right]\times\left[\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\right]\)
\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2006}+...+\frac{1}{2492}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{8.11}+\frac{9}{11.14}+...+\frac{9}{197.200}\right]\)
\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+..+\frac{1}{197}-\frac{1}{200}\right)\right]\)
\(=\left[\frac{2000}{6}\cdot\frac{498}{4996000}\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{200}\right)\right]\)
\(=\frac{83}{2498}\times\left[\frac{9}{3}\cdot\frac{3}{25}\right]\)
\(=\frac{83}{2498}\times\frac{9}{25}=\frac{747}{62450}\)
Tính tổng: \(B=\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\)
\(B=\frac{9}{8\cdot11}+\frac{9}{11\cdot14}+...+\frac{9}{197\cdot200}\)
\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\left(\frac{24}{200}-\frac{1}{200}\right)\)
\(=3\cdot\frac{23}{200}\)
đúng
\(\Rightarrow B=3\left(\frac{3}{8.11}\right)+3\left(\frac{3}{11.14}\right)+..+3\left(\frac{3}{197.200}\right)\)
\(\Rightarrow B=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)
\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{200}\right)=3.\frac{3}{25}=\frac{9}{25}\)
Vậy \(B=\frac{9}{25}\)
Chúc bn học tốt..!
F=32/8.11 +32/ 11.14 + 32/ 14.17+...+ 32/197.200
E 1/25.27+1/27.29+1/29.31 +...+1/73.75
G =15/90.94+15/94.98 +15/98.102 +....+15/146.150
H=10/56+10/140 +10/260+....+10/1400
\(\frac{5}{1.4}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}
\)
\(\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\)
lúc đầu ý bn là 5/1.3 đúng k, mk chỉnh lại như thế cho tiện nhé
a) \(\frac{5}{1\times3}+\frac{5}{3\times5}+\frac{5}{5\times7}+...+\frac{5}{99\times101}\)
\(=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}\times\frac{100}{101}=\frac{250}{101}\)
b) \(\frac{3^2}{8\times11}+\frac{3^2}{11\times14}+\frac{3^2}{14\times17}+...+\frac{3^2}{197\times200}\)
\(=\frac{9}{8\times11}+\frac{9}{11\times14}+\frac{9}{14\times17}+...+\frac{9}{197\times200}\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\times\frac{3}{25}=\frac{9}{25}\)
Ta có \(\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)
\(\Rightarrow3^2.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)\)
\(\Rightarrow9.\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(\Rightarrow3.\left(1-\frac{1}{200}\right)\)
\(\Rightarrow3.\frac{199}{200}=\frac{597}{200}\)
a) \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
\(\frac{3^2}{8.11}\) +\(\frac{3^2}{11.14}\)+\(\frac{3^2}{14.17}\)+...+\(\frac{3^2}{197.200}\)
Đặt \(A=\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\)
\(\Leftrightarrow A=\frac{9}{8.11}+\frac{9}{11.14}+\frac{9}{14.17}+...+\frac{9}{197.200}\)
\(\Leftrightarrow\frac{1}{3}A=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)
\(\Leftrightarrow\frac{1}{3}A=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{2}{17}+...+\frac{1}{197}-\frac{1}{200}\)b
\(\Leftrightarrow\frac{1}{3}A=\frac{1}{8}-\frac{1}{200}\)
\(\Leftrightarrow\frac{1}{3}A=\frac{24}{200}\)
\(\Leftrightarrow A=\frac{24}{200}\times3\)
\(\Leftrightarrow A=\frac{72}{200}=\frac{9}{25}\)
\(=\frac{3.3}{8.11}+\frac{3.3}{11.14}+...+\frac{3.3}{197.200}\)
\(=3(\frac{3}{8.11}+\frac{3}{11.14}+..+\frac{3}{197.200})\)
\(=3(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200})\)
\(=3(\frac{1}{8}-\frac{1}{200})\)
\(=3(\frac{200}{1600}-\frac{8}{1600})\)
\(=3.\frac{192}{1600}\)
\(=\frac{576}{1600}\)
Tính
A= \(\frac{3^2}{8.11}\)+ \(\frac{3^2}{11.14}\)+ \(\frac{3^2}{14.17}\)+........+ \(\frac{3^2}{197.200}\)
Các bạn giải ra giùm mik nha ! Thanks
A=\(\frac{3.3}{8.11}\)+\(\frac{3.3}{11.14}\)+\(\frac{3.3}{14.17}\)+........+\(\frac{3.3}{197.200}\)
A=3\(\frac{3}{8.11}\)+3\(\frac{3}{11.14}\)+3\(\frac{3}{14.17}\)+............+3\(\frac{3}{197.200}\)
A=3.(\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+..............+\(\frac{3}{197.200}\))
A=3.(\(\frac{1}{8}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{17}\)+.........+\(\frac{1}{197}\)-\(\frac{1}{200}\))
A=3.(\(\frac{1}{8}\)-\(\frac{1}{200}\))
A=3.(\(\frac{50}{400}\)-\(\frac{2}{200}\))
A=3.\(\frac{48}{400}\)
A=3.\(\frac{3}{25}\)
A=\(\frac{9}{25}\)
\(\frac{7x}{2.5}+\frac{7x}{5.8}+\frac{7x}{8.11}+\frac{7x}{11.14}+\frac{7x}{14.17}+\frac{7x}{17.20}=\frac{21}{10}\)
tìm x
\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)
\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)
\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)
\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)
=> \(x=2\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)
\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)
\(=>x=1\)
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)
\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
mk đầu tiên đó
\(=1\div3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\frac{9}{20}\)
\(=\frac{3}{20}\)
B = \(\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)
\(B=\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)
\(B=3.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)\)
\(B=3.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(B=3.\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(B=3.\frac{3}{25}\)
\(\Rightarrow B=\frac{9}{25}\)
\(B=\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}.\)
\(=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)
\(=3\left(\frac{11-8}{8.11}+\frac{14-11}{11.14}+...+\frac{200-197}{197.200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\cdot\frac{3}{25}\)
\(=\frac{9}{25}\)