Question

Tìm x 

1/3x5+1/5x7+...+1/x(x+2)=1/9

 

Answer 1

Đặt B =\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{x\left(x+2\right)}\)

\(\Rightarrow2B=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{x\left(x+x+2\right)}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{x+2}\)

Vì B= \(\frac{1}{9}\)\(\Rightarrow2B=\frac{1}{9}\cdot2=\frac{2}{9}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{2}{9}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{2}{9}=\frac{3}{9}-\frac{2}{9}=\frac{1}{9}\)

\(\Rightarrow x+2=9\)

\(\Rightarrow x=9-2=7\)

Vậy x=7