Question

方程组x^2+y^2-2x-2y-8=0 x^2-3xy+y^2+4=0

Answer 1

(1) x^2+y^2-2x-2y-8=0
(2) x^2-3xy+y^2+4=0
(1a) x^2+y^2=2x+2y+8 说明：由(1)得
(3) 2x+2y+8 - 3xy +4 =0 将(1a)代入(2)
得 (2-3x) y + 2x +12 = 0, y=(2x+12)/(3x-2
(4) 由1得 (x-1)^2+(y-1)^2=10
代入上面的y，得
(x-1)^2 + [ (2x+12)/(3x-2) - 1] ^2 = 10
两边乘(3x-2)^2，得 (x-1)^2*(3x-2)^2+(x-14)^2 = 10*(3x-2)^2
展开整理得 9x^4-30x^3-52x^2+72x+160=0
分解因子 (x-2)*(x-4)*(9x^2+24x-80)=0
解方程得
X1= 2 Y1=4
X2=4 Y2=2
X3=-4/3+2/3i Y3=-4/3-2/3i
X4=-4/3-2/3i Y4=-4/3+2/3i