Question

Tìm x biết

\(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.11}\) + \(\dfrac{1}{11.4}\) + ... + \(\dfrac{1}{x.\left(x+3\right)}\) = \(\dfrac{101}{1540}\)

Answer 1

Giải:

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{15}-\dfrac{1}{3\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3x+9}=\dfrac{1}{924}\)

\(\Leftrightarrow3x+9=924\)

\(\Leftrightarrow3x=915\)

\(\Leftrightarrow x=305\)

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