Có tất cả bao nhiêu bộ ba số thực (x; y; z) thoả mãn đồng thời các điều kiện dưới đây $2^{\sqrt[3]{x^2}}{.4}^{\sqrt[3]{y^2}}{.16}^{\sqrt[3]{z^2}}=128$ và ${\left(x{y}^2+z^4\right)}^2=4+{\left(x{y}^2-z^4\right)}^2$

Chọn A

● ${\left(x{y}^2+z^4\right)}^2=4+{\left(x{y}^2-z^4\right)}^2\iff {\left(x{y}^2+z^4\right)}^2-{\left(x{y}^2-z^4\right)}^2=4$

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

$\iff 2z^4.2x{y}^2=4\iff x{y}^2.z^4=1$ (1)

● $2^{\sqrt[3]{x^2}}{.4}^{\sqrt[3]{y^2}}{.16}^{\sqrt[3]{z^2}}=128$

$$\begin{gathered} \iff 2^{\sqrt[3]{x^2}+2\sqrt[3]{y^2}+4\sqrt[3]{z^2}}=2^7 \\ \iff \sqrt[3]{x^2}+2\sqrt[3]{y^2}+4\sqrt[3]{z^2}=7 \\ \iff 7\ge 7\sqrt[7]{\sqrt[3]{x^2.y^4.z^8}} \\ \iff x^2.y^4.z^8\le 1 \end{gathered}$$

Dấu ''='' xảy ra $\iff \sqrt[3]{x^2}=\sqrt[3]{y^2}=\sqrt[3]{z^2}=1\iff x^2=y{}^2=z^2=1$ (2)

Từ (1) và (2) suy ra:$\iff \left\{\begin{array}{l}{x}^2=y{}^2=z^2=1\\ x.y^2.z^4=1\end{array}\right.\left(*\right)$

Suy ra $x.x^2.x^4=1\iff x^7=1\iff x=1$

$\begin{array}{l}\left(*\right)\iff \left\{\begin{array}{l}x=1\\ y^2=1\\ z^2=1\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ \left[\begin{array}{l}y=1\\ y=-1\end{array}\right.\\ \left[\begin{array}{l}z=1\\ z=-1\end{array}\right.\end{array}\right.\iff \left\{\begin{array}{l}\left\{\begin{array}{l}x=1\\ y=1\\ z=1\end{array}\right.\\ \left\{\begin{array}{l}x=1\\ y=1\\ z=-1\end{array}\right.\\ \left\{\begin{array}{l}x=1\\ y=-1\\ z=1\end{array}\right.\\ \left\{\begin{array}{l}x=1\\ y=-1\\ z=-1\end{array}\right.\end{array}\right.\\ \\ \end{array}$

<=> có 4 bộ (x; y; z) thoả yêu cầu bài toán.