Đặt: \(\dfrac{x}{4}=\dfrac{y}{5}=k\)
\(\Rightarrow\dfrac{x}{4}=k\Rightarrow x=4k\)
\(\Rightarrow\dfrac{y}{5}=k\Rightarrow y=5k\)
Mà: \(2x^2+y^2=228\)
Thay: \(x=4k,y=5k\) vào ta có:
\(2\cdot\left(4k\right)^2+\left(5k\right)^2=228\)
\(\Rightarrow2\cdot16k^2+25k^2=228\)
\(\Rightarrow57k^2=228\)
\(\Rightarrow k^2=228:57\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k^2=2^2\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
TH1: Khi \(k=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=2\Rightarrow x=4\cdot2=8\\\dfrac{y}{5}=2\Rightarrow y=5\cdot2=10\end{matrix}\right.\)
TH2: Khi: \(k=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=-2\Rightarrow x=4\cdot-2=-8\\\dfrac{y}{5}=-2\Rightarrow y=-2\cdot5=-10\end{matrix}\right.\)
Vậy các cặp (x;y) thỏa mãn là (8;10); (-8;-10)
Đặt x/4=y/5=k
=>x=4k;y=5k
2x^2+y^2=228
=>2*16k^2+25k^2=228
=>k^2=4
TH1: k=2
=>x=8;y=10
TH2: k=-2
=>x=-8; y=-10
