Chọn phương án A.

\(\begin{aligned}
4^{x+\tfrac{1}{2}}-5\cdot2^x+2=0\Leftrightarrow&\,2^{2\left(x+\tfrac{1}{2}\right)}-5\cdot2^x+2=0\\
\Leftrightarrow&\,2^{2x+1}-5\cdot2^x+2=0\\
\Leftrightarrow&\,2\cdot\left(2^x\right)^2-5\cdot2^x+2=0\\
\Leftrightarrow&\left[\begin{array}{l}2^x=2=2^1\\ 2^x=\dfrac{1}{2}=2^{-1}\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=1\\ x=-1.\end{array}\right.
\end{aligned}\)

Vậy \(S=\{-1;1\}\).