`3:`

`a)`s

`{(\sqrt(2)x-\sqrt(3)y=-1),((1+\sqrt(3))x-\sqrt(2)y=\sqrt(2)):}`

`{(\sqrt(2)*(\sqrt(2)x-\sqrt(3)y)=-1*\sqrt(2)),(\sqrt(3)*[(1+\sqrt(3))x-\sqrt(2)y]=\sqrt(2)*\sqrt(3)):}`

`{(2x-\sqrt(6)y=-\sqrt(2)),((\sqrt(3)+3)x-\sqrt(6)y=\sqrt(6)):}`

Trừ hai vế phương trình ta được:

`2x-(\sqrt(3)+3)x=-\sqrt(2)-\sqrt(6)`

`(-1-\sqrt(3))x=\sqrt(2)(-1-\sqrt(3))`

`x=(\sqrt(2)(-1-\sqrt(3)))/(-1-\sqrt(3))`

`x=\sqrt(2)`

Thay `x=\sqrt(2)` vào `\sqrt(2)x-\sqrt(3)y=-1` ta được:

`\sqrt(2)*\sqrt(2)-\sqrt(3)y=-1`

`2-\sqrt(3)y=-1`

`\sqrt(3)y=3`

`y=\sqrt(3)`

Vậy `(x;y)=(\sqrt(2);\sqrt(3))`

`b)`

`{(4x-3y=-10),(x/2+(5y)/4=2):}`

`{(4x-3y=-10),(8(x/2+(5y)/4)=2*8):}`

`{(4x-3y=-10),(4x+10y=16):}`

Trừ hai vế phương trình ta được:

`-3y-10y=-10-16`

`-13y=-26`

`y=2`

Thay `y=2` vào `4x-3y=-10` ta được:

`4x-3*2=-10`

`4x-6=-10`

`4x=-4`

`x=-1`

Vậy `(x;y)=(-1;2)`