Lời giải:
\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+3}=\frac{9}{256}\)
\(\Leftrightarrow \left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^x.\left(\frac{1}{2}\right)^3=\frac{9}{256}\)
\(\Leftrightarrow \left(\frac{1}{2}\right)^x.(1+\frac{1}{8})=\frac{9}{256}\Rightarrow \left(\frac{1}{2}\right)^x=\frac{1}{32}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow x=5\)