Đặt \(\frac{x}{4}=\frac{y}{5}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=4k\\y=5k\end{cases}}\)
Mà \(xy=20\)\(\Leftrightarrow4k.5k=20\)
\(\Leftrightarrow20k^2=20\)
\(\Leftrightarrow k^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)
+) Với \(k=1\Leftrightarrow\hept{\begin{cases}x=4\\y=5\end{cases}}\)
+) Với \(k=-1\Leftrightarrow\hept{\begin{cases}x=-4\\y=-5\end{cases}}\)
Vậy ...
\(\frac{x}{4}=\frac{y}{5}\Rightarrow\frac{x^2}{4^2}=\frac{x}{4}.\frac{x}{4}=\frac{x}{4}.\frac{y}{5}=\frac{xy}{20}=\frac{20}{20}=1\)
\(\Rightarrow x^2=4^2=16\Rightarrow x=\pm4\)
Với x=4 thì \(y=\frac{4}{4}.5=5\)
Với x=-4 thì \(y=\frac{-4}{4}.5=-5\)
Có \(\frac{x}{4}=\frac{y}{5}\left(x.y=20\right)\)
\(\Rightarrow\hept{\begin{cases}x=4k\\y=5k\end{cases}}\left(k=0\right)\)
\(\Rightarrow4k.5k=x.y=20\)
\(\Rightarrow20k^2=20\)
\(\Rightarrow k^2=1\)
\(\Rightarrow k\in\left\{1;-1\right\}\)
+ TH1:
Nếu k = 1 thì:
x = 4
y = 5
+ TH2:
Nếu k = -1 thì:
x = -4
y = -5
Vậy (x;y) \(\in\left\{\left(4;5\right);\left(-4;-5\right)\right\}\)
