Solution
= Solution
The <controlled-NOT gate> sends $|0x\rangle$ to $|0x\rangle$ and $|1\bar x\rangle$ to $|1x\rangle$. Hence
$$
\operatorname{CX}|\beta_{zx}\rangle
=\frac{|0\rangle+(-1)^z|1\rangle}{\sqrt2}\otimes|x\rangle
=H|z\rangle\otimes|x\rangle.
$$
Since the <Hadamard gate> satisfies $H^2=I$,
$$
\boxed{(H\otimes I)\operatorname{CX}|\beta_{zx}\rangle=|zx\rangle}.
$$