Question

Tìm y biết:

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{y\left(y+3\right)}=\frac{98}{1545}\)

Answer 1

\(\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{8}+...+\frac{1}{y}-\frac{1}{y+3}\right)=\frac{98}{1545}\)

\(\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{y+3}\right)=\frac{98}{1545}\)

\(\frac{1}{5}-\frac{1}{y+3}=\frac{98}{1545}:\frac{1}{3}=\frac{98}{515}\)

\(\frac{1}{y+3}=\frac{1}{103}\)

\(y+3=103\)

\(y=100\)

Answer 2

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+......+\frac{1}{y}-\frac{1}{\left(y+3\right)}\right)=\frac{98}{1545}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{98}{1545}\)

\(\Rightarrow\left(\frac{1}{5}-\frac{1}{y+3}\right)=\frac{98}{1545}:\frac{1}{3}\)