Giải các phương trình sau
1/$\left\{\begin{matrix} x^{2}+2y^{2}-3x+2xy=0\\xy(x+y)+(x-1)^{2}=3y(1-y) \\ \end{matrix}\right.$
2/$\left\{\begin{matrix} x^{3}+3xy^{2}=-49\\x^{2}-8xy+y^{2}=8y-17x \\ \end{matrix}\right.$
3/$\left\{\begin{matrix} 6x^{2}y+2y^{3}+35=0\\5x^{2}+5y^{2}+2xy+5x+13y=0 \\ \end{matrix}\right.$
bài 1
Pt(1).y-Pt(2)$\Leftrightarrow 2y^3+xy^2-3xy-x^2-3y^2+2x+3y-1=0$
$\Leftrightarrow (2y+x-1)(y^2-y-x+1)=0$
$\left\{\begin{matrix} 2y^2+x^2-3x+2xy=0(3) \\ y^2-y-x+1=0(4) \end{matrix}\right.$
pt(3)-2pt(4)$\Leftrightarrow x^2+2xy-x+2y-2=0$
$\Leftrightarrow (x+1)(2y+x-2)=0$
bài 2
PT(1)+3PT(2)$\Leftrightarrow x^3+3xy^2+49+3x^2-24xy-24y+3y^2+51x=0$
$\Leftrightarrow(x^3+3x^3+3x+1)+48(x+1)-24y(x+1)+3y^2(x+1)=0$
$\Leftrightarrow(x+1)\left [(x+1)^2+48-24y+3y^2 \right ]=0$
$\Leftrightarrow(x+1)\left [(x+1)^2+3(y-4)^2 \right ]=0$
bài 3
PT(1)+3PT(2)$\Leftrightarrow 2y^3+6x^2y+35+15x^2+15y^2+6xy+15x+39y=0$
$\Leftrightarrow 2(y^3+3.\frac{5}{2}y^2+3.\frac{25}{4}y+\frac{125}{8})+(6x^2y+15x^2)+(6xy+15x)+\frac{3}{2}(y+\frac{5}{2})=0$
$\Leftrightarrow 2(y+\frac{5}{2})^3+6x^2(y+\frac{5}{2}+6x(y+\frac{5}{2})+\frac{3}{2}(y+\frac{5}{2})=0$
$\Leftrightarrow (y+\frac{5}{2})\left [2(y+\frac{5}{2})^2+6x^2+6x+\frac{3}{2} \right ]=0$