Đặt k = \(\dfrac{x}{4}=\dfrac{y}{7}\Rightarrow x=4k,y=7k\)
Từ x.y = 112, ta có: 4k.7k = 112
\(\Rightarrow\) \(28k^2\) = 112
\(\Rightarrow k^2=4\)
\(\Rightarrow\left[{}\begin{matrix}k=-2\\k=2\end{matrix}\right.\)
Có 2 trường hợp xảy ra:
TH1: k = -2
\(\Rightarrow x=-8,y=-14\)
TH2: k = 2
\(\Rightarrow x=8,y=14\)
Vậy \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-8\\y=-14\end{matrix}\right.\\\left\{{}\begin{matrix}x=8\\y=14\end{matrix}\right.\end{matrix}\right.\)
Vì \(\dfrac{x}{4}=\dfrac{y}{7}\)
\(\Rightarrow7.x=4.y\)
\(\Rightarrow x=\dfrac{4}{7}.y\)
Mà \(x.y=112\)
hay \(\dfrac{4}{7}.y.y=112\)
\(y^2=196\)
\(\Rightarrow\left\{{}\begin{matrix}y=14\\y=-14\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
Vậy \(y=14;x=8\)
\(y=-14;x=-8\)
Ta có:
\(\dfrac{x^2}{4^2}=\dfrac{y^2}{7^2}\)
Vì \(\dfrac{x}{4}=\dfrac{y}{7}\) nên \(\dfrac{x^2}{4^2}=\dfrac{y^2}{7^2}=\dfrac{xy}{4\cdot7}=\dfrac{112}{28}=4\)
\(\dfrac{x^2}{4^2}=4\Rightarrow x^2=4\cdot4^2=64\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
\(\dfrac{y^2}{7^2}=4\Rightarrow y^2=4\cdot7^2=196\Rightarrow\left[{}\begin{matrix}y=14\\y=-14\end{matrix}\right.\)
Vì \(xy=112\) nên ta có \(x=8;y=14\) hoặc \(x=-8;y=-14\)
