Question

已知5x^2+2x^2+2xy-14x-10y+17=0 则x= y=

Answer 1

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)

约束条件只有(x+y)^2>=0,(y-5)^2>=0

但最后一项系数为2,故可能的解为

x+y,y-5相等或相反

2x+1 与x-4相为相反数,并与上述两项绝对值相等

2x+1=-(x-4)得x=1

x+y=-(y-5) 得y=2