A = \(\frac{3^2}{8.11}\)+ \(\frac{3^2}{11.14}\) + ...... + \(\frac{3^2}{2011.2014}\)
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\(\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}\)
=\(3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)
= \(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)
= \(3\left(\frac{1}{2}-\frac{1}{17}\right)\)
=\(\frac{45}{34}\)
\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)
=3(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17)
=3(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17)
=3(1/2-1/17)
=45/34
cô Nhung ơi k đúng cho con đi cô pls
\(B=\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)
\(B=3.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)\)
\(B=3.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(B=3.\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(B=3.\frac{3}{25}\)
\(\Rightarrow B=\frac{9}{25}\)
\(B=\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}.\)
\(=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)
\(=3\left(\frac{11-8}{8.11}+\frac{14-11}{11.14}+...+\frac{200-197}{197.200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\cdot\frac{3}{25}\)
\(=\frac{9}{25}\)
Ta có :
\(x-\left(\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{1997.2000}\right)=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{1997.2000}\right)=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{1997}-\frac{1}{2000}\right)=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-3\left(\frac{1}{8}-\frac{1}{2000}\right)=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-3.\frac{249}{2000}=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-\frac{747}{2000}=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x=\frac{-1}{2}+\frac{747}{2000}\)
\(\Leftrightarrow\)\(x=\frac{-253}{2000}\)
Vậy \(x=\frac{-253}{2000}\)
Chúc bạn học tốt ~
E =\(\frac{3.3}{8.11}+\frac{3.3}{11.14}+.........+\frac{3.3}{197.200}\)
E =3.(\(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}+.....+\frac{1}{200}\))
E =3.(\(\frac{1}{8}-\frac{1}{200}\))
E =3.(\(\frac{25}{200}-\frac{1}{200}\))
E =3.\(\frac{24}{200}\)
E =3.\(\frac{3}{25}\)
E =\(\frac{3}{1}.\frac{3}{25}\)
E =\(\frac{9}{25}\)
\(B=\frac{9}{8\cdot11}+\frac{9}{11\cdot14}+...+\frac{9}{197\cdot200}\)
\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\left(\frac{24}{200}-\frac{1}{200}\right)\)
\(=3\cdot\frac{23}{200}\)
đúng
\(H=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{197.200}\right)=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{197}-\frac{1}{200}\right)=3\cdot\left(\frac{1}{2}-\frac{1}{200}\right)==\frac{297}{200}\)
\(\left[\frac{2000}{2000.2006}+\frac{2000}{2006.2012}+...+\frac{2000}{2492.2498}\right]\times\left[\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\right]\)
\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2006}+...+\frac{1}{2492}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{8.11}+\frac{9}{11.14}+...+\frac{9}{197.200}\right]\)
\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+..+\frac{1}{197}-\frac{1}{200}\right)\right]\)
\(=\left[\frac{2000}{6}\cdot\frac{498}{4996000}\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{200}\right)\right]\)
\(=\frac{83}{2498}\times\left[\frac{9}{3}\cdot\frac{3}{25}\right]\)
\(=\frac{83}{2498}\times\frac{9}{25}=\frac{747}{62450}\)
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 \(\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)
\(\Rightarrow3^2.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)\)
\(\Rightarrow9.\frac{1}{3}.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(\Rightarrow3.\left(1-\frac{1}{200}\right)\)
\(\Rightarrow3.\frac{199}{200}=\frac{597}{200}\)
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
\(\Leftrightarrow\frac{1}{3}A=\frac{1}{8}-\frac{1}{200}\)
\(\Leftrightarrow\frac{1}{3}A=\frac{24}{200}\)
\(\Leftrightarrow A=\frac{24}{200}\times3\)
\(\Leftrightarrow A=\frac{72}{200}=\frac{9}{25}\)
Ta có :
\(A=\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{2011.2014}\)
\(A=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2011.2014}\right)\)
\(A=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)
\(A=3\left(\frac{1}{8}-\frac{1}{2014}\right)\) ( khử bỏ các phân số đối nhau )
\(A=3.\frac{1003}{8056}\)
\(A=\frac{3009}{8056}\)
Vậy \(A=\frac{3009}{8056}\)
Chúc bạn học tốt ~
\(=3.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2011}-\frac{1}{2014}\right).\)
sau tự tính