用配方法化标准二次型：f(x1,x2,x3,x4)=2x1x2+2x1x3+2x1x4+2x2x3+2x2x4+2x3x4

f(x1,x2,x3,x4)=2x1x2+2x1x3+2x1x4+2x2x3+2x2x4+2x3x4
= 2(y1+y2)(y1-y2)+2(y1+y2)y3+2(y1+y2)y4+2(y1-y2)y3+2(y1-y2)y4+2y3y4
= 2y1^2 - 2y2^2 + 4y1y3 + 4y1y4 + 2y3y4
= 2(y1+y3+y4)^2 - 2y2^2 -2y3^2 -2y4^2 - 2y3y4
= 2(y1+y3+y4)^2 - 2y2^2 -2(y3+(1/2)y4)^2 -(3/2)y4^2
= 2z1^2 - 2z2^2 - 2z3^2 - (3/2)z4^2像这类问题都是令前两个构成平方差，其余不变吗是的