giải
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
vậy ta có : \(4x+\frac{15}{16}=1\)
suy ra : \(4x=1-\frac{15}{16}\)
\(4x=\frac{1}{16}\)
\(x=\frac{1}{16}:4=\frac{1}{16}x\frac{1}{4}=\frac{1}{64}\)
Vậy \(x=\frac{1}{64}\)