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\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)
\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{y\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y+3}{5\left(y+3\right)}-\dfrac{5}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y+3-5}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y-2}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow515\left(y-2\right)=98.5\left(y+3\right)\)
\(\Leftrightarrow515y-1030=490y+1470\)
\(\Leftrightarrow25y-2500=0\\ \Leftrightarrow25y=2500\\ \Leftrightarrow y=100\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{y+3}\right)=\dfrac{98}{1545}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{103}\)
hay x=100
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\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)
\(\Leftrightarrow\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{y\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{14}-...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{5}-\dfrac{98}{515}\\ \Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{103}\\ \Leftrightarrow y+3=103\\ \Leftrightarrow y=100\)
Vậy...........................................
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\(\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{8}+...+\frac{1}{y}-\frac{1}{y+3}\right)=\frac{98}{1545}\)
\(\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{y+3}\right)=\frac{98}{1545}\)
\(\frac{1}{5}-\frac{1}{y+3}=\frac{98}{1545}:\frac{1}{3}=\frac{98}{515}\)
\(\frac{1}{y+3}=\frac{1}{103}\)
\(y+3=103\)
\(y=100\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+......+\frac{1}{y}-\frac{1}{\left(y+3\right)}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\left(\frac{1}{5}-\frac{1}{y+3}\right)=\frac{98}{1545}:\frac{1}{3}\)
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tìm y đúng hơn là tính :
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{y\left(y+3\right)}=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{y\left(y+3\right)}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{y}-\frac{1}{y+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{y+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{y+3}=\frac{98}{1545}\div\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{y+3}=\frac{98}{515}\)
\(\Rightarrow\frac{1}{y+3}=\frac{1}{5}-\frac{98}{515}\)
\(\Rightarrow\frac{1}{y+3}=\frac{5}{515}=\frac{1}{103}\)
\(\Rightarrow y+3=103\)
\(\Rightarrow y=100\)
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\(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{3\cdot101}{1540}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{308}{1540}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(x+3=308\)
\(x=308-3=305\)
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\(\frac{1}{10}\) + \(\frac{1}{40}\) +\(\frac{1}{88}\)+\(\frac{1}{154}\) + \(\frac{1}{238}\) + \(\frac{1}{340}\)
= \(\frac{1}{8}\) + \(\frac{1}{56}\) + \(\frac{1}{140}\)
= \(\frac{1}{7}\) + \(\frac{1}{140}\)
= \(\frac{3}{20}\)
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\(\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+...+\dfrac{1}{y.(y+3)}=\dfrac{98}{1545}\)
\(3.(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{y(y+3)})=\dfrac{98}{1545}.3\)
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{y(y+3)}=\dfrac{98}{515}\)
\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(y+3.1=1.515\)
\(y+3=515\)
\(y=515-3\)
\(y=512\)
Vậy y = 512
Nhớ tick cho mk nha
= 1/5.8+1/8.11+1/11.14+...+1/y(y+3)= 98/1545
=1/3.(1/5-1/8+1/8-1/11+1/11-1/14+...+1/y-1/y+3)=98/1545
=1/3.(1/5-1/y+3)=98/1545
=1/5-1/y+3=98/1545:1/3=98/515
=1/1+3=1/103
y+3=103
y=103-3
y=100