Giải:
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
Ta có: \(xy=6\)
\(\Rightarrow6k^2=6\)
\(\Rightarrow k^2=1\)
\(\Rightarrow k=\pm1\)
+) \(k=1\Rightarrow x=2,y=3\)
+) \(k=-1\Rightarrow x=-2,y=-3\)
Vậy...
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
Vì xy = 6 \(\Rightarrow\) 2k . 3k = 6
6k = 6
k = 1
Vậy \(\left\{{}\begin{matrix}x=2.1=2\\y=3.1=3\end{matrix}\right.\)
Xét: \(\dfrac{x}{2}=\dfrac{y}{3}=k\)
\(\Rightarrow x=2k;y=3k\)
\(\Rightarrow xy=2k.3k=6k^2\)
\(\Rightarrow6k^2=6\Rightarrow k^2=1\Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)
TH1: \(x=-2;y=-3\) (loại)
TH2: \(x=2;y=3\) (chọn)
Vậy x=2; y=3
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\) ta có \(x=2k\), \(y=3k\)
Ta có : x.y = 2k . 3k = 6
\(\Rightarrow\) \(6k^2\) = 6
\(\Rightarrow\) \(k^2\) = 1
\(\Rightarrow\) k = 1
\(\Rightarrow\) x = 1.2 = 2; y = 1.3 = 3 nha bạn
\(\text{Ta có:}\) \(\dfrac{x}{2}=\dfrac{y}{3}\)
\(\Leftrightarrow\left(\dfrac{x}{2}\right)^2=\dfrac{x}{2}\cdot\dfrac{y}{3}\)
\(\Leftrightarrow\left(\dfrac{x}{2}\right)^2=\dfrac{xy}{6}\)
\(\Leftrightarrow\left(\dfrac{x}{2}\right)^2=\dfrac{6}{6}\)
\(\Leftrightarrow\left(\dfrac{x}{2}\right)^2=1\)
\(\Leftrightarrow\dfrac{x}{2}=1\)
\(\Leftrightarrow x=2\)
\(\text{Mà}\) \(xy=6\)
\(\Leftrightarrow2y=6\)
\(\Leftrightarrow y=3\)
\(\text{Vậy}\) \(x=2\\ y=3\)
