Question

Tìm x 

1/2×4 + 1/4×6 + ....+ 1/[(2x-2) × 2x = 1/8

Answer 1

\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left[\left(2x-2\right).2x\right]}=\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)=\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{8}:\frac{1}{2}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}\)

\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)