Question

[Nguyễn Ngọc Gia Hân]

Tìm x;y;z biết :

x.y =\(\dfrac{1}{2}\) ; y.z=\(\dfrac{3}{5}\) ; z.x = \(\dfrac{27}{10}\)

Answer 1

[Luân Đào]

\(\left\{{}\begin{matrix}xy=\dfrac{1}{2}\\yz=\dfrac{3}{5}\\zx=\dfrac{27}{10}\end{matrix}\right.\Rightarrow xyyzzx=\dfrac{1}{2}\cdot\dfrac{3}{5}\cdot\dfrac{27}{10}\Leftrightarrow\left(xyz\right)^2=\dfrac{81}{100}\)

\(\Rightarrow\left[{}\begin{matrix}xyz=-\dfrac{9}{10}\\xyz=\dfrac{9}{10}\end{matrix}\right.\)

+ Khi \(xyz=-\dfrac{9}{10}\)

\(\Rightarrow\left\{{}\begin{matrix}z=-\dfrac{9}{10}:\dfrac{1}{2}=-\dfrac{9}{5}\\x=-\dfrac{9}{10}:\dfrac{3}{5}=-\dfrac{3}{2}\\y=-\dfrac{9}{10}:\dfrac{27}{10}=-\dfrac{1}{3}\end{matrix}\right.\)

+ Khi \(xyz=\dfrac{9}{10}\)

\(\Rightarrow\left\{{}\begin{matrix}z=\dfrac{9}{10}:\dfrac{1}{2}=\dfrac{9}{5}\\x=\dfrac{9}{10}:\dfrac{3}{5}=\dfrac{3}{2}\\y=\dfrac{9}{10}:\dfrac{27}{10}=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(\dfrac{3}{2};\dfrac{1}{3};\dfrac{9}{5}\right);\left(-\dfrac{3}{2};-\dfrac{1}{3};-\dfrac{9}{5}\right)\)

Answer 2

[Ngô Tấn Đạt]

\(\left(x.y\right).\left(y.z\right)\left(z.x\right)=\dfrac{1}{2}.\dfrac{3}{5}.\dfrac{27}{10}\\ \Rightarrow\left(x.y.z\right)^2=\dfrac{81}{100}\\ \Rightarrow\left[{}\begin{matrix}x.y.z=\dfrac{9}{10}\\x.y.z=-\dfrac{9}{10}\end{matrix}\right.\)

Nếu x.y.z=9/10

\(\Rightarrow z=\dfrac{9}{10}:\dfrac{1}{2}=\dfrac{9}{5};x=\dfrac{9}{10}:\dfrac{3}{5}=\dfrac{3}{2};y=\dfrac{9}{10}:\dfrac{27}{10}=\dfrac{1}{3}\)

Nếu x.y.z=-9/10

\(\Rightarrow z=-\dfrac{9}{5};x=-\dfrac{3}{2};y=-\dfrac{1}{3}\)