Đặt \(\dfrac{x}{4}=\dfrac{y}{6}=k\)
=> \(\left\{{}\begin{matrix}x=4.k\\y=6.k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x.y=24.k^2\\dox.y=24\end{matrix}\right.\left\{{}\begin{matrix}\Rightarrow24.k^2=24\\k^2=24.24\\k^2=1\\\Rightarrow k=\pm1\end{matrix}\right.\)
Với : k=1 => \(\left\{{}\begin{matrix}x=4.1=4\\y=6.1=6\end{matrix}\right.\)
Với : k = -1 => \(\left\{{}\begin{matrix}x=4.\left(-1\right)=-4\\y=6.\left(-1\right)=6\end{matrix}\right.\)
Kết luận : x,y = ( 4;6), ( -4 ; -6 )
Có: \(\dfrac{x}{4}=\dfrac{y}{6}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{xy}{6}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{24}{6}\)
\(\Rightarrow\dfrac{x^2}{4}=4\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
Vậy \(x\in\left\{-4;4\right\}\)
Đặt \(\dfrac{x}{4}=\dfrac{y}{6}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=6k\end{matrix}\right.\)
\(xy=24\)
\(\Leftrightarrow4k.6k=24\)
\(k=-1\)
\(\Leftrightarrow24k^2=24\)
\(\Leftrightarrow k^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}k=1\\k=-1\end{matrix}\right.\)
+) \(k=1\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)
+) \(k=-1\) \(\Leftrightarrow\left\{{}\begin{matrix}k=-4\\k=-6\end{matrix}\right.\)
Vậy..
Làm một hoài chán lắm=> Phải thêm một cách giải khác mới ok!
\(\dfrac{x}{4}=\dfrac{y}{6}\Leftrightarrow6x=4y\Leftrightarrow6.4=xy\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-4\\y=-6\end{matrix}\right.\)
Vậy ....
Ta có: \(\dfrac{x}{4}=\dfrac{y}{6}\\ 6x=4y\\ =>4xy=6x^2\)
=>4xy=6x2=24
=>x2=4
=>x=2
Mà xy=24
=>y=12
Áp dụng dãy tỉ số bằng nhau :
\(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{x\times y}{4\times6}=\dfrac{24}{24}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=1\rightarrow x=4\\\dfrac{y}{6}=1\rightarrow y=6\end{matrix}\right.\)
Vậy : x\(=\)4 , y\(=\)6
