a: Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
Ta có: xy=6
nên \(6k^2=6\)
hay \(k^2=1\)
Trường hợp 1: k=1
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\\y=3k=3\end{matrix}\right.\)
Trường hợp 2: k=-1
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-2\\y=3k=-3\end{matrix}\right.\)
b: Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)
Ta có: xy=20
nên \(20k^2=20\)
hay \(k^2=1\)
Trường hợp 1: k=1
\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=4\\y=5k=5\end{matrix}\right.\)
Trường hợp 2: k=-1
\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=-4\\y=5k=-5\end{matrix}\right.\)
a. Đặt \(\dfrac{x}{2}=\dfrac{y}{3}\) = k
\(\dfrac{x}{2}=k\) => x = 2k
\(\dfrac{y}{3}=k\) => y = 3k
2k * 3k = 6 <=> 6k^2 = 6
<=> k^2 = 6 : 6
<=> k^2 = 1
<=> k = +- 1 (mik ko bik dấu cộng trừ ở đâu hết -.-)
\(\dfrac{x}{2}\)= +- 1 <=> x = 1 * +- 2 = +- 2
\(\dfrac{x}{3}\)= +- 1 <=> x = 1 * +- 3 = +- 3
Vậy x = +- 2 ; y = +- 3
