Theo đề ta có:
\(\dfrac{x}{3}\)\(=\dfrac{y}{7}\) và \(\)\(x.y=84\) \(\left(m^2\right)\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{7}=k\)\(\Rightarrow x=3k\)
\(y=7k\) (*)
Thay (*) vào \(x.y=84\left(m^2\right)\) ta được:
\(3k.7k=84\)\(\Rightarrow21k^2=84\Rightarrow k^2=84:21=4\)
\(\Rightarrow k^2=\left(2\right)^2\) hoặc \(k^2=\left(-2\right)^2\)
\(\Rightarrow k=2\) hoặc \(k=-2\)
\(TH1:\)
\(k=2\Rightarrow x=3k=3.2=6\left(m\right)\)
\(y=7k=7.2=14\left(m\right)\)
\(TH2\):
\(k=-2\Rightarrow x=3k=3.\left(-2\right)=-6\left(m\right)\)
\(y=7k=7.\left(-2\right)=-14\left(m\right)\)
Vậy ...............
