1)ĐK:$\begin{cases}x\geq 0;y\geq \frac{-1}{3}\\ 2x-y-1\geq 0;x+2y\geq 0\end{cases}$(*)pt(1)$\Leftrightarrow$$\sqrt{2x-y-1}$-$\sqrt{x+2y}$+$\sqrt{3y+1}$-$\sqrt{x}$=0
$\Leftrightarrow$(x-3y-1)($\frac{1}{\sqrt{2x-y-1}+\sqrt{x+2y}}$-$\frac{1}{\sqrt{3y+1}+\sqrt{x}}$)=0
TH1:x=3y+1
Thế vào pt(2)$\Rightarrow$$(3y+1)^{3}$-$3(3y+1)+2$=2$y^{3}$-$y^{2}$
$\Leftrightarrow$$y^{2}$(25y+28)=0$\Rightarrow$y=0 (t/m(*))or y=$\frac{-28}{25}$(L)
$\Rightarrow$(x;y)=(1;0)
TH2:$\frac{1}{\sqrt{2x-y-1}+\sqrt{x+2y}}$=$\frac{1}{\sqrt{3y+1}+\sqrt{x}}$
$\Leftrightarrow$$\sqrt{2x-y-1}$-$\sqrt{x}$+$\sqrt{x+2y}$-$\sqrt{3y+1}$=0
$\Leftrightarrow$$(x-y-1)(\frac{1}{\sqrt{2x-y-1}+\sqrt{x}}+\frac{1}{\sqrt{x+2y}+\sqrt{3y+1}}$)=0
$\Leftrightarrow$x=y+1(do(...)>0)
Tương tự TH1$\Rightarrow$(x;y)=...