f=2*x1x2+2*x1x3-2*x1x4-2*x2x3+2*x2x4+2*x3x4用配方法化为标准型!

令 x1=y1+y2,x2=y1-y2,x3=y3,x4=y4
f = 2y1^2-2y2^2+4y2y3-4y2y4+2y3y4
= 2y1^2-2(y2-y3+y4)^2+2y3^2+2y4^2-2y3y4
= 2y1^2-2(y2-y3+y4)^2+2(y3-(1/2)y4)^2+(3/2)y4^2
令 z1=y1,z2=y2-y3+y4,z3=y3-(1/2)y4,z4=y4
则 f = 2z1^2-2z2^2+2z3^2+(3/2)z4^2.