Đặt: \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)
\(\Rightarrow xy=2k\cdot3k=6k^2=54\Rightarrow k^2=9\Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)
+) Với k = 3 ta có: \(\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=9\end{matrix}\right.\)
+) Với k = -3 ta có: \(\left\{{}\begin{matrix}x=2k=2\cdot\left(-3\right)=-6\\y=3k=3\cdot\left(-3\right)=-9\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(6;9\right);\left(-6;-9\right)\)
\(\dfrac{x}{2}=\dfrac{y}{3}\&xy=54\)
Áp dụng tc dãy tỉ số BN ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{xy}{2.3}=\dfrac{54}{6}=9\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=9\\\dfrac{y}{3}=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=27\end{matrix}\right.\)
Ta có:
\(\dfrac{x}{2}-\dfrac{y}{3}=\dfrac{x.y}{2.3}=\dfrac{54}{6}=9\)
x=2.3=6
y=3.3=9
\(Đặt: \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\) \(\Rightarrow xy=2k\cdot3k=6k^2=54\Rightarrow k^2=9\Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\) +) Với k = 3 ta có: \(\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=9\end{matrix}\right.\) +) Với k = -3 ta có: \(\left\{{}\begin{matrix}x=2k=2\cdot\left(-3\right)=-6\\y=3k=3\cdot\left(-3\right)=-9\end{matrix}\right.\) Vậy \(\left(x;y\right)=\left(6;9\right);\left(-6;-9\right)\)\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}\) = K
Ta có : \(x=2.K\) ; \(y=3.K\)
\(x.y=54\)
= \(2.K.3.K=54\)
= \(6.K^2=54\)
\(K^2=54:6\)
➜\(K^2=9\)
➜\(K=3\)
\(\dfrac{x}{2}=3\xrightarrow[]{}\)\(x=6\)
\(\dfrac{y}{3}=3\xrightarrow[]{}y=9\)