Có tất cả bao nhiêu bộ ba số thực $\left( x,y,z \right)$ thỏa mãn...

* ${{\left( x{{y}^{2}}+{{z}^{4}} \right)}^{2}}=4+{{\left( x{{y}^{2}}-{{z}^{4}} \right)}^{2}}\Rightarrow {{\left( x{{y}^{2}}+{{z}^{4}} \right)}^{2}}-{{\left( x{{y}^{2}}-{{z}^{4}} \right)}^{2}}=4$
$\Leftrightarrow \left( x{{y}^{2}}+{{z}^{4}}-x{{y}^{2}}+{{z}^{4}} \right)\left( x{{y}^{2}}+{{z}^{4}}+x{{y}^{2}}-{{z}^{4}} \right)=4\Rightarrow 2x{{y}^{2}}.2{{z}^{4}}=4$
$\Rightarrow x{{y}^{2}}{{z}^{4}}=1\left( x>0 \right)\left( 1 \right)$
* ${{2}^{\sqrt[3]{{{x}^{2}}}}}{{.4}^{\sqrt[3]{{{y}^{2}}}}}{{.16}^{\sqrt[3]{{{z}^{2}}}}}=128\Rightarrow {{2}^{\sqrt[3]{{{x}^{2}}}+2\sqrt[3]{{{y}^{2}}}+4\sqrt[3]{{{z}^{2}}}}}={{2}^{7}}$
$\Rightarrow \sqrt[3]{{{x}^{2}}}+2\sqrt[3]{{{y}^{2}}}+4\sqrt[3]{{{z}^{2}}}=7\left( 2 \right)$
Đặt $\sqrt[3]{x}=a;\sqrt[3]{y}=b;\sqrt[3]{z}=c$
$\Rightarrow $ Hệ phương trình $\left\{ \begin{aligned}
& {{a}^{2}}+2{{b}^{2}}+4{{c}^{2}}=7 \\
& {{a}^{3}}.{{b}^{6}}.{{c}^{12}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+2{{b}^{2}}+4{{c}^{2}}=7 \\
& a.{{b}^{2}}.{{c}^{4}}=1 \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& {{a}^{2}}+2.\dfrac{1}{a.{{c}^{4}}}+4{{c}^{2}}=7\left( * \right) \\
& {{b}^{2}}=\dfrac{1}{a{{c}^{4}}} \\
\end{aligned} \right.$
Vế trái phương trình $\left( * \right):{{a}^{2}}+\dfrac{1}{a{{c}^{4}}}+\dfrac{1}{a{{c}^{4}}}+{{c}^{2}}+{{c}^{2}}+{{c}^{2}}+{{c}^{2}}\ge 7.\sqrt[7]{{{a}^{2}}.\dfrac{1}{a{{c}^{4}}}.\dfrac{1}{a{{c}^{4}}}.{{c}^{2}}.{{c}^{2}}.{{c}^{2}}.{{c}^{2}}}$
$\Rightarrow VT\ge 7.$ Để dấu "=" xảy ra $\Rightarrow \left\{ \begin{aligned}
& {{a}^{2}}={{c}^{2}} \\
& {{a}^{2}}=\dfrac{1}{a{{c}^{4}}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& {{b}^{2}}=1 \\
& {{c}^{2}}=1 \\
\end{aligned} \right.$
$\Rightarrow $ 4 cặp $\left( a,b,c \right)\Rightarrow $ 4 bộ $\left( x,y,z \right).$