Sửa lại đề theo bạn ns:
Ta có:
\(xy.yz.xz=\dfrac{1}{2}.\dfrac{3}{5}.\dfrac{27}{10}\)
\(\Rightarrow\left(xyz\right)^2=\dfrac{81}{100}\Rightarrow xyz=\pm\dfrac{9}{10}\)
Xét \(xyz=-\dfrac{9}{10}\) ta có:
\(\left\{{}\begin{matrix}x=xyz:yz=-\dfrac{9}{10}:\dfrac{3}{5}=-\dfrac{3}{2}\\y=xyz:xz=-\dfrac{9}{10}:\dfrac{27}{10}=-\dfrac{1}{3}\\z=xyz:xy=-\dfrac{9}{10}:\dfrac{1}{2}=-\dfrac{9}{5}\end{matrix}\right.\).
Xét \(xyz=\dfrac{9}{10}\) ta có:
\(\left\{{}\begin{matrix}x=xyz:yz=\dfrac{9}{10}:\dfrac{3}{5}=\dfrac{3}{2}\\y=xyz:xz=\dfrac{9}{10}:\dfrac{27}{10}=\dfrac{1}{3}\\z=xyz:xy=\dfrac{9}{10}:\dfrac{1}{2}=\dfrac{9}{5}\end{matrix}\right.\)
Vậy............ Chúc bạn học tốt!!!
Mạng vs chả lỗi @@!
Ta có:
\(2y=\dfrac{3}{5}\Rightarrow y=\dfrac{3}{10}\)
Thay \(y=\dfrac{3}{10}\) vào \(xy=\dfrac{1}{2}\) ta được:
\(\dfrac{3}{10}x=\dfrac{1}{2}\Rightarrow x=\dfrac{5}{3}\)
Thay \(x=\dfrac{5}{3}\) vào \(xz=\dfrac{27}{10}\) ta được:
\(\dfrac{5}{3}z=\dfrac{27}{10}\Rightarrow z=\dfrac{81}{50}\)
Chúc bạn học tốt!!!
Ta có: \(y.2=\dfrac{3}{5}\Rightarrow y=\dfrac{3}{10}\)
\(x.y=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{2}:y=\dfrac{1}{2}.\dfrac{10}{3}=\dfrac{5}{3}\)
\(z.x=\dfrac{27}{10}\Rightarrow z=\dfrac{27}{10}:x=\dfrac{17}{10}.\dfrac{3}{5}=\dfrac{51}{50}\)
Vậy.............
