Đặt \(\frac{x}{2}=\frac{y}{3}=k\left(k\inℚ\right)\)
=>\(\hept{\begin{cases}x=2k\\y=3k\end{cases}}\)
ta có xy=54
(=) 2k.3k=54
(=) \(6.k^2\)=54
(=) \(k^2=9\)
=> k=3
=> \(\hept{\begin{cases}x=2.3\\y=3.3\end{cases}\left(=\right)\hept{\begin{cases}x=6\\y=9\end{cases}}}\)
Đặt : \(\frac{x}{2}=\frac{y}{3}=k\)
\(\Rightarrow x=2k;y=3k\)
Khi đó : \(2k.3k=54\)
\(\Rightarrow6k^2=54\)
\(\Rightarrow k^2=54:6=9=3^2\)
\(\Rightarrow k=3\)hoặc \(k=-3\)
\(\Rightarrow x=2.3=6\)\(;y=3.3=9\)hoặc
\(x=2.\left(-3\right)=-6\)\(;y=3.\left(-3\right)=-9\)
Ta có :
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{54}{2.3}=\frac{54}{6}=9\)
\(\Rightarrow\frac{x}{2}.\frac{y}{3}=\left(\frac{x}{2}\right)^2=\left(\frac{y}{3}\right)^2\)
\(\Rightarrow\left(\frac{x}{2}\right)^2=\left(\frac{y}{3}\right)^2=9\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}=3\\\frac{x}{2}=\frac{y}{3}=-3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=3.2;y=3.3\\x=-3\cdot2;y=-3.3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=6;y=9\\x=-6;y=-9\end{cases}}\)
Vậy ....
Đặt \(\frac{x}{2}=\frac{y}{3}=k\)
\(\Rightarrow x=2.k;\)\(y=3.k\)
Mà xy= 54
\(\Rightarrow\)2k.3k= 54
\(k^2.6=54\)
\(k^2=54:6\)
\(k^2=9\)
\(\Rightarrow k^2=3^2\)hoặc \(k^2=\left(-3\right)^2\)
k= 3. k= -3
- Với k= 3 \(\Rightarrow\orbr{\begin{cases}x=2.3\Rightarrow x=6\\y=3.3\Rightarrow y=9\end{cases}}\)
- Với k= -3 \(\Rightarrow\hept{\begin{cases}x=2.\left(-3\right)=-6\\y=3.\left(-3\right)=-9\end{cases}}\)
Vậy x= 6 và y= 9 hoặc x= -6 và y= -9
Đặt x = 2k ; y = 3k
xy=54
2k.3k=54
6k^2=54
k^2=9
\(k=\pm3\)
\(\Rightarrow\hept{\begin{cases}x=6\\y=9\end{cases},\hept{\begin{cases}x=-6\\y=-9\end{cases}}}\)
