giải hệ pt
1, $\left\{\begin{matrix}y^{2}+(4x-1)^{2}=\sqrt[3]{4x(8x+1)} & & \\ & & \\ 40x^{2}+x=y\sqrt{14x-1}& & \end{matrix}\right.$
2, $\left\{\begin{matrix}\sqrt{x}+\sqrt{4-y}=x^{2}-y-1 & \\ \sqrt{2(x-y)^{2}+6x-2y+4}-\sqrt{y}=\sqrt{x+1}& \end{matrix}\right.$
3, $\left\{\begin{matrix}(x+y)^{2}=x+y+2xy+4 & \\ xy(x+y)+8=x^{2}+y^{2}+4x\sqrt{y-1}+4y\sqrt{x-1}& \end{matrix}\right.$