Đáp án: $(x,y)\in\{(0,0), (1,\dfrac23)\}$
Giải thích các bước giải:
Ta có:
$\dfrac{x+y}{5}=\dfrac{x-y}{1}=\dfrac{xy}{2}=\dfrac{x+y+x-y}{5+1}=\dfrac{2x}{6}=\dfrac{x}{3}$
$\to \dfrac{xy}{2}=\dfrac{x}{3}$
$\to 3xy=2x$
$\to 3xy-2x=0$
$\to x(3y-2)=0$
$\to x=0$ hoặc $3y-2=0\to 3y=2\to y=\dfrac23$
Nếu $x=0$
$\to \dfrac{0+y}{5}=\dfrac{0-y}{1}\to \dfrac{y}{5}=\dfrac{-y}{1}$
$\to -y=5y$
$\to 6y=0$
$\to y=0$
$\Rightarrow (x,y)=(0,0)$
Nếu $y=\dfrac23$
$\to \dfrac{x+\dfrac23}{5}=\dfrac{x-\dfrac23}{1}=\dfrac{x\cdot\dfrac23}{2}$
$\to \dfrac{x+\dfrac23}{5}=\dfrac{x-\dfrac23}{1}$
$\to x+\dfrac23=5(x-\dfrac23)$
$\to x+\dfrac23=5x-\dfrac{10}{3}$
$\to 5x-x=\dfrac{10}{3}+\dfrac{2}{3}$
$\to 4x=\dfrac{12}{3}=4$
$\to x=1$
