Vì:
`\qquad x(x-y)=1/9\ne 0`
`\qquad y(x-y)=-1/ 3\ne 0`
`=>x\ne 0;y\ne 0;x-y\ne 0=>x\ne y`
$\\$
`\qquad x(x-y)=1/9`
`=> -3x(x-y)=-3. 1/ 9`
`=> -3x(x-y)=-1/ 3`
Mà ` y(x-y)=-1/ 3`
`=> -3x(x-y)=y(x-y)`
Vì `x-y\ne 0`
`=>{-3x(x-y)}/{x-y}={y(x-y)}/{x-y}`
`=>-3x=y`
$\\$
Thay `y=-3x` vào `x(x-y)=1/ 9` ta có:
`\qquad x(x+3x)=1/ 9`
`=>x.4x=1/ 9`
`=>x^2=1/{36}`
$⇒\left[\begin{array}{l}x=\dfrac{1}{6}\\x=\dfrac{-1}{6}\end{array}\right.$ $⇒\left[\begin{array}{l}y=-3x=-3.\dfrac{1}{6}=\dfrac{-1}{2}\\y=-3x=-3.\dfrac{-1}{6}=\dfrac{1}{2}\end{array}\right.$
$\\$
Vậy các cặp số `(x;y)` thỏa đề bài là:
`(1/ 6;{-1}/2);({-1}/6;1/ 2)`
