Vì \(\dfrac{x}{2}=\dfrac{y}{5}\Rightarrow x=2k;y=5k\) (1)
Thay \(x\cdot y=10\) vào (1), ta có:
\(2k\cdot5k=10\)
\(\Rightarrow10k^2=10\)
\(\Rightarrow k^2=1\)
\(\Rightarrow k=\pm1\)
Nếu \(k=1\) thì: \(\left\{{}\begin{matrix}x=2\cdot1=2\\y=5\cdot1=5\end{matrix}\right.\)
Nếu \(k=-1\) thì \(\left\{{}\begin{matrix}x=2\cdot\left(-1\right)=-2\\y=5\cdot\left(-1\right)=-5\end{matrix}\right.\)
Đặt :
\(\dfrac{x}{2}=\dfrac{y}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)
Thay \(\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\) vào \(x.y=10\) ta được :
\(x.y=2k.5k=10\)
\(\Leftrightarrow10k^2=10\)
\(\Leftrightarrow k^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}k^2=1^2\\k^2=\left(-1\right)^2\end{matrix}\right.\)
+) \(k=1\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)
+) \(k=-1\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-5\end{matrix}\right.\)
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