Giải thích các bước giải:
$\left\{\begin{matrix}2(x+y)-\sqrt{x-1}=12\\x+y+3\sqrt{x-1}=13\end{matrix}\right.$`=>`$\left\{\begin{matrix}2(x+y)-\sqrt{x-1}=12\\2(x+y)+6\sqrt{x-1}=26\end{matrix}\right.$
`=>6sqrt(x-1)+sqrt(x-1)=26-12`
`=>7sqrt(x-1)=14`
`=>sqrt(x-1)=2`
`=>x-1=4=>x=5`
$\left\{\begin{matrix}2(x+y)-\sqrt{x-1}=12\\x+y+3\sqrt{x-1}=13\end{matrix}\right.$`=>`$\left\{\begin{matrix}6(x+y)-3\sqrt{x-1}=36\\x+y+3\sqrt{x-1}=13\end{matrix}\right.$
`=>6(x+y)+x+y=36+13`
`=>7(x+y)=49`
`=>x+y=7`
Mà `x=5=>y=2`
Vậy `x=5;y=2.`
