`a)` `x+y-2\sqrt{x-1}-4\sqrt{y-3}+1=0 \ (x\ge 1, y\ge 3)`
`<=>(x-1)-2\sqrt{x-1}+1+(y-3)-4\sqrt{y-3}+4=0`
`<=>` `(\sqrt{x-1}-1)^2+(\sqrt{y-3}-2)^2=0`
Do $\begin{cases}(\sqrt{x-1}-1)^2\ge 0 \\ (\sqrt{y-3}-2)^2 \ge 0 \\\end{cases}$
`=>`$\begin{cases}(\sqrt{x-1}-1)^2=0\\(\sqrt{y-3}-2)^2 =0\\\end{cases}$
`<=>`$\begin{cases}\sqrt{x-1}=1\\ \sqrt{y-3}=2\\\end{cases}$`<=>`$\begin{cases}x-1=1\\y-3=4\\\end{cases}$
`<=>`$\begin{cases}x=2\ (tm) \\y=7\ (tm)\\\end{cases}$
Vậy phương trình có nghiệm: `(x;y)=(2;7)`
`b)` `x+y+z-2\sqrt{x-2}+2\sqrt{y+2006}-2\sqrt{x-2007}=0 \ (x\ge 2\, \ y\leq -2006\, z\ge 2007)`
`<=>(x-2)-2\sqrt{x-1}+1+(y+2006)+2\sqrt{y+2006}+1+(z-2007)-2\sqrt{x-2007}+1=0`
`<=> (\sqrt{x-2}-1)^2+(\sqrt{y+2006}+1)^2+(\sqrt{x-2007}-1)^2=0`
Do : $\begin{cases} (\sqrt{x-2}-1)^2\ge 0\\ (\sqrt{y+2006}+1)^2\ge 0\\ (\sqrt{x-2007}-1)^2\ge 0\\\end{cases}$
`=>`$\begin{cases} (\sqrt{x-2}-1)^2=0\\ (\sqrt{y+2006}+1)^2=0\\ (\sqrt{x-2007}-1)^2=0\\\end{cases}$
`<=>`$\begin{cases}\sqrt{x-2}=1\\\sqrt{y+2006}=-1 (ktm)\\\sqrt{x-2007}=1\\\end{cases}$
`<=>`$\begin{cases}x=3(tm)\\y=\sqrt{y+2006}=-1(ktm)\\z=2008(tm)\\\end{cases}$
`=>` Phương trình vô nghiệm.
