\(\dfrac{3}{2.5}\)+\(\dfrac{3}{5.8}\)+\(\dfrac{3}{8.11}\)\(\dfrac{3}{11.14}\)+\(\dfrac{3}{14.17}\)<\(\dfrac{1}{2}\)
\(\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+\dfrac{3^2}{14.17}+....+\dfrac{3^2}{197.200}\)
\(\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+...+\dfrac{3^2}{197.200}\)
=\(3.\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{197.200}\right)\)
=\(3.\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
=\(3.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
=\(3.\dfrac{3}{25}=\dfrac{9}{25}\)
Ta có: \(\dfrac{3^2}{8\cdot11}+\dfrac{3^2}{11\cdot14}+\dfrac{3^2}{14\cdot17}+...+\dfrac{3^2}{197\cdot200}\)
\(=3\left(\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+...+\dfrac{3}{197\cdot200}\right)\)
\(=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(=3\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)
Tìm x biết: \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+\dfrac{3x}{8.11}+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{3}{2.5}\) \(+
\) \(\dfrac{3}{5.8}\) \(+\) \(\dfrac{3}{8.11}\) \(+\) \(\dfrac{3}{11.14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{5}\) \(+\) \(\dfrac{1}{5}\) \(-\) \(\dfrac{1}{8}\) \(+\) \(\dfrac{1}{8}\) \(-\) \(\dfrac{1}{11}\) \(+\) \(\dfrac{1}{11}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) x \(\dfrac{3}{7}\) \(=\) \(\dfrac{1}{21}\)
\(x\) \(=\) \(\dfrac{1}{21}\) \(:\) \(\dfrac{3}{7}\)
\(x\) \(=\) \(\dfrac{1}{9}\)
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)
\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{y\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y+3}{5\left(y+3\right)}-\dfrac{5}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y+3-5}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y-2}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow515\left(y-2\right)=98.5\left(y+3\right)\)
\(\Leftrightarrow515y-1030=490y+1470\)
\(\Leftrightarrow25y-2500=0\\ \Leftrightarrow25y=2500\\ \Leftrightarrow y=100\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{y+3}\right)=\dfrac{98}{1545}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{103}\)
hay x=100
\(Tìmxbiết\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x+\left(x+3\right)}\)
Ta có: \(\dfrac{k}{x.\left(x+k\right)}=\dfrac{x+k-x}{x.\left(x+k\right)}=\dfrac{1}{x}-\dfrac{1}{x+k}\)
nên áp dụng ta có:
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\)
\(=\dfrac{1}{5}-\dfrac{1}{x+3}\)
Nên $\dfrac{1}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right)=\dfrac{1}{3}.(\dfrac{1}{5}-\dfrac{1}{x+3})$
Đến đây là làm được rồi nha
So sánh:
A = \(\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+\dfrac{3^2}{8.11}\) và B = \(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
So sánh: A =\(\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+\dfrac{3^2}{8.11}\) và B \(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
vậy \(x=305\)
Tìm x:
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+.......+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
Ta có : \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+....+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
= \(\dfrac{1}{3}\) . ( \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+....-\dfrac{1}{x+3}\)
=\(\dfrac{1}{3}\). ( \(\dfrac{1}{5}-\dfrac{1}{x+3}\)) = \(\dfrac{101}{1540}\)
=>\(\dfrac{1}{5}-\dfrac{1}{x+3}\) = \(\dfrac{303}{1540}\)
=> \(\dfrac{1}{x+3}\)= \(\dfrac{5}{1540}=\dfrac{1}{308}\)
=> x+3 = 308
=> x= 305
Vậy x= 305
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x-3}\right)=\dfrac{101}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x-3}=\dfrac{101}{1540}:\dfrac{1}{3}\)
Tìm x :
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\) ( x # 0 ; x# - 3)
⇔ \(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
⇔ \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
⇔ \(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
⇔ \(\dfrac{1}{x+3}=\dfrac{1}{308}\)
⇔ \(x+3=308\)
⇔ \(x=305\left(TM\right)\)
Vậy ,...