$\quad \begin{cases}2x+y=5m-1\ (1)\\x-2y=2\ (2)\end{cases}$
`(2)=>x=2y+2` thay vào $(1)$
`(1)<=>2(2y+2)+y=5m-1`
`<=>4y+4+y=5m-1`
`<=>5y=5m-5`
`<=>y=m-1`
`\qquad x=2y+2=2(m-1)+2`
`<=>x=2m`
Theo đề bài:
`\qquad x^2-2y^2=1`
`<=>(2m)^2-2(m-1)^2=1`
`<=>4m^2-2(m^2-2m+1)-1=0`
`<=>4m^2-2m^2+4m-2-1=0`
`<=>2m^2+4m-3=0`
`<=>2(m^2+2m+1)-5=0`
`<=>2(m+1)^2=5⇔(m+1)^2=5/ 2`
$⇔\left[\begin{array}{l}m+1=\sqrt{\dfrac{5}{2}}\\m+1=-\sqrt{\dfrac{5}{2}}\end{array}\right.$
$⇔\left[\begin{array}{l}m=-1+\sqrt{\dfrac{5}{2}}\\m=-1-\sqrt{\dfrac{5}{2}}\end{array}\right.$
Vậy `m\in {-1+\sqrt{5/ 2}; -1-\sqrt{5/2}}`
