Question

tìm x biết 

a 1/5.8+1/8.11+1/11.14+...1/x(x+3)=101/1540

b 1+1/3+1/6+1/10+...1/x(x+1)=\(1\frac{1991}{1993}\)

Answer 1

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

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

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x.3}=\frac{303}{1540}\)

\(=\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(=\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(=\frac{1}{x+3}=\frac{1}{308}\)

\(x+3=308\)

\(\Rightarrow x=305\)