1/5*8 + 1/8*11 + 1/11*14 + 1/x*(x+3) = 101/1540
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\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=305\)
Pikachu đơn giản thì làm thử đừng nói mà ko làm nha ^_^
duyệt đi
\(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{5}+\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\left(\frac{1}{5}-\frac{1}{x+3}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{x}-\frac{1}{x}\right)=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=>x+3=308
x=308-3
x=305
Vậy x=305
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1504}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\Rightarrow\frac{x-2}{5x+15}=\frac{303}{1540}\)
\(\Rightarrow x=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+30+\frac{101}{1540}\right)}\)
Phải gì không .
a)\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
b ko hiểu đề
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{308}{1540}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{5}{1540}=\frac{1}{308}\)
=> x + 3 = 308
=> x = 308 - 3
=> x = 305
Vậy x = 305
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{5}{1540}=\frac{1}{308}\)
=> x + 3 = 308
=> x = 308 - 3
=> x = 305
Vậy x = 305