$\left\{\begin{matrix} (x+\sqrt{x^{2}+1})(y+\sqrt{y^{2}+1})=1 & & \\ 3x^{2}+y+3=\sqrt{3x+1}+\sqrt{4-5y}& & \end{matrix}\right.$
Đk: $x\geq \frac{-1}{3},y\leq 0.8$
Từ (1). Mà $(y+\sqrt{y^{2}+1})(\sqrt{y^{2}+1}-y)=1$, $(\sqrt{x^{2}+1}+x)(\sqrt{x^{2}+1}-x)=1$
$\Rightarrow x=-y$ thay vào (2) tđ:
$3x^{2}-x+3=\sqrt{3x+1}+\sqrt{4+5x}$
$\Leftrightarrow 3x(x-1)=\left [ \sqrt{3x+1} -(x+1)\right ]+\left [ \sqrt{4+5x}-(x+2) \right ]$
$\Leftrightarrow 3x(x-1)=\frac{x(1-x)}{\sqrt{3x+1}+x+1}+\frac{x(1-x)}{\sqrt{4+5x}+x+2}$
Mà $3+\frac{1}{\sqrt{3x+1}+x+1}+\frac{1}{\sqrt{4+5x}+x+2}> 0\vee x\geq \frac{-1}{3}$
$\Rightarrow x(x-1)=0$
N x=0(t/m) $\Rightarrow$ y=0(t/m)
N x=1(t/m) $\Rightarrow$ y=-1(t/m)