\(\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+...+\dfrac{3^2}{197.200}\)

=\(3.\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{197.200}\right)\)

=\(3.\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

=\(3.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)

=\(3.\dfrac{3}{25}=\dfrac{9}{25}\)

Ta có: \(\dfrac{3^2}{8\cdot11}+\dfrac{3^2}{11\cdot14}+\dfrac{3^2}{14\cdot17}+...+\dfrac{3^2}{197\cdot200}\)

\(=3\left(\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+...+\dfrac{3}{197\cdot200}\right)\)

\(=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

\(=3\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)

\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)

\(B=\frac{9}{8\cdot11}+\frac{9}{11\cdot14}+...+\frac{9}{197\cdot200}\)

\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(=3\left(\frac{24}{200}-\frac{1}{200}\right)\)

\(=3\cdot\frac{23}{200}\)

đúng

Đặt 3 ra ngoài

\(\Rightarrow B=3\left(\frac{3}{8.11}\right)+3\left(\frac{3}{11.14}\right)+..+3\left(\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{200}\right)=3.\frac{3}{25}=\frac{9}{25}\)

Vậy \(B=\frac{9}{25}\)

Chúc bn học tốt..!

=1,798

Ta có : 

\(C=\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\)

\(C=3\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\right)\)

\(C=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(C=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(C=3.\frac{3}{25}\)

\(C=\frac{9}{25}\)

Chúc bạn học tốt ~ 

kết quả bài nãy bằng kkkkk

N=3(3/8.11 +3 /11.14 + 3/14.17 +...+3/197.200)

N=3( 1/8-1/11+1/11-1/14+1/14-1/17+...+1/197-1/200)

N=3(1/8-1/200)

N=3. 3/25=9/25

Ủng hộ mk nha

\(\Rightarrow N=\frac{9}{8.11}+\frac{9}{11.14}+\frac{9}{14.17}+....+\frac{9}{197.200}\)

\(\Rightarrow N=\frac{9}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(\Rightarrow N=3\left(\frac{1}{8}-\frac{1}{200}\right)=\frac{3}{8}-\frac{3}{200}=\frac{75}{200}-\frac{3}{200}=\frac{72}{200}=\frac{9}{25}\)

N = 3²/8.11  +  3²/11.14  +  3²/14.17  + ...+3²/197.200.

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

\(\Leftrightarrow\frac{1}{3}A=\frac{1}{8}-\frac{1}{200}\)

\(\Leftrightarrow\frac{1}{3}A=\frac{24}{200}\)

\(\Leftrightarrow A=\frac{24}{200}\times3\)

\(\Leftrightarrow A=\frac{72}{200}=\frac{9}{25}\)

Thank

\(=\frac{3.3}{8.11}+\frac{3.3}{11.14}+...+\frac{3.3}{197.200}\)

\(=3(\frac{3}{8.11}+\frac{3}{11.14}+..+\frac{3}{197.200})\)

\(=3(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200})\)

\(=3(\frac{1}{8}-\frac{1}{200})\)

\(=3(\frac{200}{1600}-\frac{8}{1600})\)

\(=3.\frac{192}{1600}\)

\(=\frac{576}{1600}\)

\(\left[\frac{2000}{2000.2006}+\frac{2000}{2006.2012}+...+\frac{2000}{2492.2498}\right]\times\left[\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\right]\)

\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2006}+...+\frac{1}{2492}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{8.11}+\frac{9}{11.14}+...+\frac{9}{197.200}\right]\)

\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+..+\frac{1}{197}-\frac{1}{200}\right)\right]\)

\(=\left[\frac{2000}{6}\cdot\frac{498}{4996000}\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{200}\right)\right]\)

\(=\frac{83}{2498}\times\left[\frac{9}{3}\cdot\frac{3}{25}\right]\)

\(=\frac{83}{2498}\times\frac{9}{25}=\frac{747}{62450}\)

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) \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\frac{100}{101}\)

\(=\frac{250}{101}\)