Giải các hệ phương trình sau (có thể dùng phương pháp đưa về tích số)1,\left\{\begin{matrix} y^2+x\sqrt{\frac{2(y^2+3)}{x}}=3(4x-1) & \\ \sqrt[3]{y^2-7x+27}+\sqrt{12-x}=2(8x-y^2) & \end{matrix}\right.2,\left\{\begin{matrix} 17(x-y)=3xy-2x^2-y^2\\ \sqrt{x+3}+\sqrt{10-y}=x^2-7y+11 \end{matrix}\right.3,\left\{\begin{matrix} 2(\frac{y+1}{x^3}+xy)+\frac{y^2+2y+7}x=(3x^2-2)\\2\sqrt{x-1}+3x\sqrt{-4-y}=x^2y+72 \end{matrix}\right.5,\left\{\begin{matrix} x+y+6=2\sqrt{(2y-x)(x+4)}\\ 2(x\sqrt{5x-y+3}-5)=\frac{12(y-x)}{x^3+4}-\sqrt{y-4} \end{matrix}\right.6,\left\{\begin{matrix} x^3-3y^3=x^2y+5xy^2\\ 2(\sqrt{3x}+\sqrt{2y-1})-7=11y+6\sqrt{y(x-y-1)} \end{matrix}\right.8,\begin{Bmatrix} 5(3-\sqrt{5x+y})=2x-\frac{3y}{x}\\ \sqrt{2x^3-29}+\sqrt[3]{x^2+2x-9+y}=\frac{91-y-10x}{x+2} \end{Bmatrix}9,\left\{\begin{matrix} x^2+2x^2y-3xy^2+x(y+1)=2y^2(5y+1)+2y\\ (x^2+17y+12)^2=4(x+y+7)(x^2+3x+8y+5) \end{matrix}\right.10,\left\{\begin{matrix} 2x^3+(5+y)x^2+y^2(2x+5)+2y(5x+2)=-y^3-2x\\x^2+y^2+2x+5y+2=0 \end{matrix}\right.11,\left\{\begin{matrix} 2\sqrt{x^2+5x-y+2}-2=\sqrt{y^2+8x}+x\\ 2y-\sqrt{x+3} =x+11 \end{matrix}\right.