Tìm x,y,z thõa mãn : $\dfrac{x}{2}$ = $\dfrac{y}{3}$ = $\dfrac{z}{1}$ Và $x^{2}$ . $y^{2}$ = $z^{2}$ Trong T/C dãy tỉ số bằng nhau ko có nhân đâu

$\begin{array}{l}\text{Đáp án:}\\\text{Đặt:} \dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{1}=k\\\to \begin{cases}x=2k\\y=3k\\z=k\end{cases} \ \ (1)\\\text{Thay (1) vào $x^2\times y^2=z^2$, ta có:}\\(2k)^2\times (3k)^2=k^2\\\to 4k^2\times 9k^2=k^2\\\to 36k^4=k^2\\\to 36k^2=1\\\to \left[\begin{array}{l}k=-\dfrac{1}{6}\\k=\dfrac{1}{6}\end{array}\right.\\\text{Với $k=-\dfrac{1}{6}$, ta có:} \\ \begin{cases}x=-\dfrac{1}{6}\times2\\y=-\dfrac{1}{6}\times3\\z=-\dfrac{1}{6}\times1\end{cases} \ \to \begin{cases}x=-\dfrac{1}{3}\\y=-\dfrac{1}{2}\\z=-\dfrac{1}{6}\end{cases}\\\text{Với $k=\dfrac{1}{6}$, ta có:}\\\begin{cases}x=\dfrac{1}{6}\times2\\y=\dfrac{1}{6}\times3\\z=\dfrac{1}{6}\times1\end{cases} \ \to \begin{cases}x=\dfrac{1}{3}\\y=\dfrac{1}{2}\\z=\dfrac{1}{6}\end{cases}\\\text{Vậy $(x;y;z)=(-\dfrac{1}{3};-\dfrac{1}{2};-\dfrac{1}{6}),(\dfrac{1}{3};\dfrac{1}{2};\dfrac{1}{6})$}\end{array}$