Chọn A.

* \({{\left( x{{y}^{2}}+{{z}^{4}} \right)}^{2}}=4+{{\left( x{{y}^{2}}-{{z}^{4}} \right)}^{2}}\)\( \Rightarrow {{\left( x{{y}^{2}}+{{z}^{4}} \right)}^{2}}-{{\left( x{{y}^{2}}-{{z}^{4}} \right)}^{2}}=4\)

\(\Leftrightarrow \left( x{{y}^{2}}+{{z}^{4}}-x{{y}^{2}}+{{z}^{4}} \right)\left( x{{y}^{2}}+{{z}^{4}}+x{{y}^{2}}-{{z}^{4}} \right)=4\)\( \Rightarrow 2x{{y}^{2}}.2{{z}^{4}}=4\)

\(\Rightarrow x{{y}^{2}}{{z}^{4}}=1\text{ }\left( x>0 \right)\left( 1 \right)\)

* \({{2}^{\sqrt[3]{{{x}^{2}}}}}{{.4}^{\sqrt[3]{{{y}^{2}}}}}{{.16}^{\sqrt[3]{{{z}^{2}}}}}=128\)\( \Rightarrow {{2}^{\sqrt[3]{{{x}^{2}}}+2\sqrt[3]{{{y}^{2}}}+4\sqrt[3]{{{z}^{2}}}}}={{2}^{7}}\)

\(\Rightarrow \sqrt[3]{{{x}^{2}}}+2\sqrt[3]{{{y}^{2}}}+4\sqrt[3]{{{z}^{2}}}=7\text{ }\left( 2 \right)\)

Đặt \(\sqrt[3]{x}=a;\sqrt[3]{y}=b;\sqrt[3]{z}=c\)

\(\Rightarrow \) Hệ phương trình

\(\begin{array}{l}
\left\{ \begin{array}{l}
{a^2} + 2{b^2} + 4{c^2} = 7\\
{a^3}.{b^6}.{c^{12}} = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{a^2} + 2{b^2} + 4{c^2} = 7\\
a.{b^2}.{c^4} = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{a^2} + 2.\frac{1}{{a.{c^4}}} + 4{c^2} = 7{\rm{ }}\left( * \right)\\
{b^2} = \frac{1}{{a{c^4}}}
\end{array} \right.
\end{array}\)

Vế trái phương trình \(\left( * \right):{{a}^{2}}+\frac{1}{a{{c}^{4}}}+\frac{1}{a{{c}^{4}}}+{{c}^{2}}+{{c}^{2}}+{{c}^{2}}\)\( +{{c}^{2}}\ge 7.\sqrt[7]{{{a}^{2}}.\frac{1}{a{{c}^{4}}}.\frac{1}{a{{c}^{4}}}.{{c}^{2}}.{{c}^{2}}.{{c}^{2}}.{{c}^{2}}}\)

\(\Rightarrow VT\ge 7.\) Để dấu “=” xảy ra 

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} {a^2} = {c^2}\\ {a^2} = \frac{1}{{a{c^4}}} \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} a = 1\\ {b^2} = 1\\ {c^2} = 1 \end{array} \right. \end{array}\)

\(\Rightarrow \) 4 cặp \(\left( a,b,c \right)\Rightarrow \) 4 bộ \(\left( x,y,z \right).\)