`ĐKXĐ: x, y ne 0, xy ne 0`
\(\left\{\begin{matrix}\dfrac{4x + 5y}{xy} = 2\\20x - 30y + xy = 0\end{matrix}\right.\)
`<=>`\(\left\{\begin{matrix}\dfrac{4x + 5y}{xy} = \dfrac{2xy}{xy}\\2(20x - 30y + xy) = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}4x + 5y = 2xy\\40x - 60y + 2xy = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}-4x - 5y + 2xy = 0\\40x - 60y + 2xy = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}40x - 60y + 2xy - 44x + 55y = 0\\4x - 60y + 2xy = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}-44x + 55y = 0\\40x - 60y + 2xy = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}y = \dfrac{4}{5}x\\40x - 60. (\dfrac{4}{5}x) + 2x. (\dfrac{4}{5}x) = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}y = \dfrac{4}{5}x\\40x - 48x + \dfrac{8}{5}x^2 = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}y = \dfrac{4}{5}x\\ -8x + \dfrac{8}{5}x^2 = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}y = \dfrac{4}{5}x\\8x(-1 + \dfrac{1}{5}x) = 0\end{matrix}\right.\)
Ta có: `8x(-1 + 1/5x) = 0`
`<=>` \(\left[ \begin{array}{l}x=0\\x = 5\end{array} \right.\)
`+) x = 0 <=> y = 4/5. 0 = 0`(Trái với điều kiện đề bài, Loại)
`+) x = 5 <=> y = 4/5. 5 = 4`(Thoả mãn, Chọn)
Vậy `x = 5, y = 4`
