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ND
0
TM
27 tháng 2 2017
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<=>\(\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)
<=>\(\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{202}{1540}\)
<=>\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{202}{1540}\)
<=>\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{202}{1540}:\frac{2}{3}=\frac{303}{1540}\)
<=>\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
<=> x+3=308
<=> x=305
9 tháng 2 2017
\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+..+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)
\(\Leftrightarrow\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)
\(\Leftrightarrow\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{202}{1540}\)
\(\Leftrightarrow\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{202}{1540}:\frac{2}{3}=\frac{303}{1540}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Rightarrow x=305\)
Vạy x = 305
28 tháng 2 2017
\(\dfrac{2}{40}+\dfrac{2}{88}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{202}{1540}\)
\(\Leftrightarrow2\left(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\right)=\dfrac{202}{1540}\)
\(\Leftrightarrow\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
BT
1
\(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{3\cdot101}{1540}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{308}{1540}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(x+3=308\)
\(x=308-3=305\)