5x^2+2y^2+2xy-14x-10y+17
=(x^2+2xy+y^2)+(4x^2-14x-8)+ (y^2-10y+25)
=(x+y)^2+(y-5)^2+2(2x+1)(x-4)
約束條件只有(x+y)^2>=0,(y-5)^2>=0
但最后一項(xiàng)系數(shù)為2,故可能的解為
x+y,y-5相等或相反
2x+1 與x-4相為相反數(shù),并與上述兩項(xiàng)絕對值相等
2x+1=-(x-4)得x=1
x+y=-(y-5) 得y=2