a, Ta có: \(\left\{{}\begin{matrix}xy=\dfrac{1}{3}\\yz=\dfrac{-2}{5}\\xz=\dfrac{-3}{10}\end{matrix}\right.\Rightarrow x^2y^2z^2=\dfrac{1}{25}\Rightarrow xyz=\pm\dfrac{1}{5}\)
+) Xét \(xyz=\dfrac{1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}xyz:xy=\dfrac{1}{5}:\dfrac{1}{3}\\xyz:yz=\dfrac{1}{5}:\dfrac{-2}{5}\\xyz:xz=\dfrac{1}{5}:\dfrac{-3}{10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=\dfrac{3}{5}\\x=\dfrac{-1}{2}\\y=\dfrac{-2}{3}\end{matrix}\right.\)
+) Xét \(xyz=\dfrac{-1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}z=\dfrac{-3}{5}\\x=\dfrac{1}{2}\\y=\dfrac{-2}{3}\end{matrix}\right.\)
Vậy....
b, Ta có: \(\left\{{}\begin{matrix}x+y=\dfrac{-7}{6}\\y+z=\dfrac{1}{4}\\x+z=\dfrac{1}{12}\end{matrix}\right.\Rightarrow2\left(x+y+z\right)=-\dfrac{5}{6}\)
\(\Rightarrow x+y+z=\dfrac{-5}{12}\)
\(\Rightarrow\left\{{}\begin{matrix}z=\dfrac{-5}{12}+\dfrac{7}{6}\\x=\dfrac{-5}{12}-\dfrac{1}{4}\\y=-\dfrac{5}{12}-\dfrac{1}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=\dfrac{3}{4}\\x=\dfrac{-2}{3}\\y=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy...
a) Ta có: \(\left(x.y\right)\left(y.z\right)\left(x.z\right)=\dfrac{1}{3}.\dfrac{-2}{5}.\dfrac{-3}{10}=\dfrac{1}{25}\)
\(\left(xyz\right)^2=\dfrac{1}{25}\Rightarrow\left[{}\begin{matrix}xyz=\dfrac{1}{5}\\xyz=-\dfrac{1}{5}\end{matrix}\right.\)
*\(xyz=\dfrac{1}{5}\)
\(x=\dfrac{xyz}{yz}=\dfrac{1}{5}:\dfrac{-2}{5}=-\dfrac{1}{2}\)
\(y=\dfrac{xy}{x}=\dfrac{1}{3}:\dfrac{-1}{2}=-\dfrac{2}{3}\)
\(z=\dfrac{xz}{x}=\dfrac{-3}{10}:\dfrac{-1}{2}=\dfrac{3}{5}\)
*\(xyz=-\dfrac{1}{5}\)
\(x=\dfrac{xyz}{yz}=\dfrac{-1}{5}:\dfrac{-2}{5}=\dfrac{1}{2}\)
\(y=\dfrac{xy}{x}=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\)
\(z=\dfrac{xz}{x}=\dfrac{-3}{10}:\dfrac{1}{2}=-\dfrac{3}{5}\)
Vậy \(\left(x,y,z\right)=\dfrac{-1}{2};\dfrac{-2}{3};\dfrac{3}{5}\) hoặc \(\left(x,y,z\right)=\dfrac{1}{2};\dfrac{2}{3};\dfrac{-3}{5}\)
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