giải hệ phương trình $\left \{ {{(x+20)(y-1)=xy} \atop {(x-10)(y+1)=xy}} \right.$

Đáp án:

Vậy `(x;y)=(40;3)`

Giải thích các bước giải:

`{((x+20)(y-1)=xy),((x-10)(y+1)=xy):}`

`<=>{(xy-x+20y-20=xy),(xy+x-10y-10=xy):}`

`<=>{(xy-x+20y-20-xy=0),(xy+x-10y-10-xy=0):}`

`<=>{(-x+20y-20=0),(x-10y-10=0):}`

`<=>{(x=20y-20),(20y-20-10y-10=0):}`

`<=>{(x=20y-20),(10y-30=0):}`

`<=>{(x=20y-20),(y=3):}`

`<=>{(x=20.3-20),(y=3):}`

`<=>{(x=40),(y=3):}`

Vậy `(x;y)=(40;3)`