Question

Tìm x biết:

\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1504}\)

Dấu ''.'' là dấu nhân 

Answer 1

\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1504}\)

\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1504}\)

\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1504}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1504}:\frac{1}{3}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1504}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1504}\)

\(\frac{1}{x+3}=-\frac{11}{7502}\)

\(x+3=\left(7502.1\right):\left(-11\right)\)

\(x+3=7502:\left(-11\right)\)

\(x+3=-682\)

\(x=-682-3\)

\(x=-385\)

Answer 2

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