cách 1:\(\dfrac{2}{x}=\dfrac{3}{y}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{xy}{2y}=\dfrac{96}{2y}\)
Ta có: \(\dfrac{y}{3}=\dfrac{96}{2y}\Rightarrow2y^2=288\Leftrightarrow y^2=144\Leftrightarrow\left[{}\begin{matrix}y=12\Rightarrow x=8\\y=-12\Rightarrow x=-8\end{matrix}\right.\)
Vậy (x;y) = (8;12) ; (-8;-12)
cách 2: \(\dfrac{2}{x}=\dfrac{3}{y}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)
Đặt: \(k=\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow x=2k;y=3k\)
\(\Rightarrow xy=2k\cdot3k=6k^2\)
hay 96 = 6k2
=> k2 = 16 \(\Leftrightarrow\left[{}\begin{matrix}k=4\\k=-4\end{matrix}\right.\)
+) Với k = 4 => \(\left\{{}\begin{matrix}x=2\cdot4=8\\y=3\cdot4=12\end{matrix}\right.\)
+) Với k = -4 =>\(\left\{{}\begin{matrix}x=2\cdot\left(-4\right)=-8\\y=3\cdot\left(-4\right)=-12\end{matrix}\right.\)
Vậy........
p/s: làm cách nào cx đc nhé
\(\dfrac{2}{x}=\dfrac{3}{y}\)và \(x.y=96\)
\(\Rightarrow\dfrac{y}{3}=\dfrac{x}{2}\)
Ta có: \(\dfrac{y}{3}=\dfrac{x}{2}=k\)
\(\Rightarrow\left\{{}\begin{matrix}y=3.k\\x=2.k\end{matrix}\right.\)mà \(x.y=96\)
\(\Rightarrow3k.2k=96\)
\(6.k^2=96\)
\(k^2=96\div6\)
\(k^2=16\)
\(\Rightarrow\)\(\)\(k=4\) hoặc \(k=-4\)
\(\Rightarrow\left\{{}\begin{matrix}y=4.3\\x=4.2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=12\\x=8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=\left(-4\right).3\\x=\left(-4\right).2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-12\\x=-8\end{matrix}\right.\)
Vậy \(y=12\) ; \(x=8\) hoặc \(y=-12\) ; \(x=-8\)