\(\frac{x+y}{5}=\frac{x-y}{1}\)
=>\(\frac{x}{5}+\frac{y}{5}=x-y\)
=>\(\frac{y}{5}+y=x-\frac{x}{5}\)
=>\(\frac{y}{5}+\frac{5y}{5}=\frac{5x}{5}-\frac{x}{5}\)
=>\(\frac{y+5y}{5}=\frac{5x-x}{5}\)
=>\(\frac{6y}{5}=\frac{4x}{5}\)
=>6y=4x
=>\(y=\frac{4}{6}.x\)
Lại có: \(\frac{x-y}{1}=\frac{x.y}{2}\)
=>2.(x-y)=x.y
=>\(2.\left(x-\frac{4}{6}.x\right)=x.y\)
=>\(2.\frac{1}{3}.x=x.y\)
=>\(\frac{2}{3}=y\)
=>\(x=\frac{2}{3}:\frac{4}{6}=1\)
Vậy x=1,\(y=\frac{2}{3}\)