\(\frac{12}{16}=\frac{-x}{4}=\frac{21}{y}=\frac{z}{80}\)
\(\Rightarrow\frac{3}{4}=\frac{-x}{4}=\frac{21}{y}=\frac{z}{80}\)
\(\frac{3}{4}=\frac{-x}{4}\Rightarrow-x=\frac{3\cdot4}{4}=3\Rightarrow x=-3\)\(\frac{-\left(-3\right)}{4}=\frac{21}{y}\Rightarrow y=\frac{21\cdot4}{-\left(-3\right)}=28\)\(\frac{21}{28}=\frac{z}{80}\Rightarrow z=\frac{21\cdot80}{28}=60\)Vậy x=-3, y=28, z=60
\(\frac{12}{16}=-\frac{x}{4}=\frac{21}{y}=\frac{z}{80}\)
\(\frac{3}{4}=-\frac{x}{4}=\frac{21}{y}=\frac{z}{80}\Rightarrow3.4=4.-x\)
\(12=4.-x\)
\(3=-x\)
\(-3=x\)Từ x=-3 suy ra ta có:
\(\frac{3}{4}=\frac{21}{y}\Rightarrow21.4=3.y\)
\(16=y\) Từ 16=y suy ra ta có:
\(\frac{21}{16}=\frac{z}{80}\Rightarrow21\cdot80=16.z\)
\(105=z\)
Vậy \(x=-3;y=16;z=105\)