\(x^4.y^4=\left(x.y\right)^4=16\Leftrightarrow x.y=2\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{4}=k\)
\(\Rightarrow\dfrac{x}{2}=k\Leftrightarrow x=2k\)
\(\Rightarrow\dfrac{y}{4}=k\Leftrightarrow y=4k\)
Mà \(x.y=2\), ta có :
\(2k.4k=2\)
\(\Leftrightarrow8k^2=2\Leftrightarrow k^2=\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}k=\dfrac{1}{2}\\k=-\dfrac{1}{2}\end{matrix}\right.\)
+) TH1: Khi \(k=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
+ ) TH2 : Khi \(k=-\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)
Vậy ......
\(x^4\times y^4=16\)
\(\Rightarrow\left(xy\right)^4=16\)
\(\Rightarrow xy=-2;2\)
Xét \(x,y=-2\)
\(\dfrac{x}{2}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{xy}{8}=-1\)
\(\Rightarrow x^2=-1\) (loại)
\(\Rightarrow xy=2\)
\(\Rightarrow x^2=1\)
\(\Rightarrow x=-1;1\)
\(x=-1;y=-2\)
\(x=1;y=2\)
Vậy \(\left(x,y\right)=\left(-1,-2\right);\left(1,2\right)\)
ta có : \(\dfrac{x}{2}=\dfrac{y}{4}\Leftrightarrow4x=2y\Leftrightarrow2x=y\)
thay vào \(x^4.y^4=16\Leftrightarrow x^4.\left(2x\right)^4=16\Leftrightarrow x^4.16x^4=16\Leftrightarrow16x^8=16\)
\(\Leftrightarrow x^8=\dfrac{16}{16}=1=1^8\Rightarrow x=1\)
ta có : \(x=1\Rightarrow y=2.1=2\) vậy \(x=1;y=2\)
\(\left(\dfrac{x}{2}\right)=\left(\dfrac{y}{4}\right)\Leftrightarrow\left(\dfrac{x^4}{16}\right)=\left(\dfrac{y^4}{256}\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^4}{16}=\dfrac{y^4}{256}=\dfrac{x^4y^4}{16.256}=\dfrac{16}{16.256}=\dfrac{1}{256}\)
\(\Rightarrow\left\{{}\begin{matrix}x^4=\dfrac{1}{256}.16=\dfrac{1}{16}\\y^4=\dfrac{1}{256}.256=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{1}{2}\\y=\pm1\end{matrix}\right.\)
