Giải:
Ta có: \(\dfrac{2}{x}=\dfrac{3}{y}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
Mà \(xy=96\)
\(\Rightarrow2k3k=96\)
\(\Rightarrow6k^2=96\)
\(\Rightarrow k^2=16\)
\(\Rightarrow\left[{}\begin{matrix}k=4\\k=-4\end{matrix}\right.\)
+) \(k=4\Rightarrow x=8,y=12\)
+) \(k=-4\Rightarrow x=-8,y=-12\)
Vậy cặp số \(\left(x;y\right)\) là: \(\left(8;12\right);\left(-8;-12\right)\)
Giải:
Ta có: \(\dfrac{2}{x}=\dfrac{3}{y}\)
\(\Rightarrow\dfrac{2}{3}=\dfrac{x}{y}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=t\)
\(\Rightarrow x=2y;y=3t\)
\(\Rightarrow x.y=96\)
\(\Rightarrow2t.3t=96\)
\(\Rightarrow6t^2=96\)
\(\Rightarrow t^2=16\)
\(\Rightarrow t=\pm4\)
Với \(t=4\) ta có:
\(x=4.2\)
\(\Rightarrow x=8\)
\(y=4.3\)
\(\Rightarrow y=12\)
Với \(t=-4\) tương tự ta cũng có:
\(x=-4.2\)
\(\Rightarrow x=-8\)
\(y=-4.3\)
\(\Rightarrow y=-12\)
Vậy .....
\(\dfrac{2}{x}=\dfrac{3}{y}\)
=> x = \(\dfrac{2}{3}y\)
Lại có: xy = 96
=> \(\dfrac{2}{3}y^2=96\)
<=> y2 = 144
=> \(\left[{}\begin{matrix}y=12\\y=-12\end{matrix}\right.\)
*) y = 12 => x = \(\dfrac{2}{3}\).12 = 8
*) y = -12 => x = \(\dfrac{2}{3}\).12 = -8
Vậy (x,y) = (8,12); (-8,-12)
Có: \(\dfrac{2}{x}=\dfrac{3}{y}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\) và \(xy=96\)
Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{xy}{2y}=\dfrac{96}{2y}\)
\(\Rightarrow\dfrac{y}{3}=\dfrac{96}{2y}\Rightarrow y\cdot2y=96\cdot3=288\)
\(\Rightarrow y^2=\dfrac{288}{2}=144\Rightarrow\left[{}\begin{matrix}y=12\\y=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{96}{12}=8\\x=\dfrac{96}{-12}=-8\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left(12;8\right);\left(-12;-8\right)\)
