Question

1/3*5+1/5*7+...+1/(2x+1)*(2x+3)=15/93

 

Answer 1

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}\right)=2.\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(2x=90\)

\(x=\frac{90}{2}=45\)

Vậy \(x=45\)