Question

1) Tìm \(x; y; z\) biết \(6\left(x - \frac{1}{y}\right) = 3\left(y - \frac{1}{z}\right) = 2\left(z - \frac{1}{x}\right) = xyz - \frac{1}{xyz}\).

Answer 1

Đặt \(6\left(x-\dfrac{1}{y}\right)=3\left(y-\dfrac{1}{z}\right)=2\left(z-\dfrac{1}{x}\right)=xyz-\dfrac{1}{xyz}=t\)

\(\Rightarrow36\left(x-\dfrac{1}{y}\right)\left(y-\dfrac{1}{z}\right)\left(z-\dfrac{1}{x}\right)=t^3\)

\(\Rightarrow36\left(xyz+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-x-y-z-\dfrac{1}{xyz}\right)=t^3\)

\(\Rightarrow36\left(xyz-\dfrac{1}{xyz}-\left(x-\dfrac{1}{y}\right)-\left(y-\dfrac{1}{z}\right)-\left(z-\dfrac{1}{x}\right)\right)=t^3\)

\(\Rightarrow36\left(t-\dfrac{t}{6}-\dfrac{t}{3}-\dfrac{t}{2}\right)=t^3\)

\(\Rightarrow t^3=0\Rightarrow t=0\)

\(\Rightarrow x-\dfrac{1}{y}=y-\dfrac{1}{z}=z-\dfrac{1}{x}=xyz-\dfrac{1}{xyz}=0\)

\(\Rightarrow\left\{{}\begin{matrix}xy=1\\yz=1\\zx=1\\\left(xyz\right)^2=1\end{matrix}\right.\)

TH1: \(xyz=1\Rightarrow x=y=z=1\)

TH2: \(xyz=-1\Rightarrow x=y=z=-1\)