f(x,y,z,w)=x+y+z+w+a(x²+y²+z²+w²+x+2y+3z+4w-17/2)
f`x=1+a(2x+1)=0
f`y=1+a(2y+2)=0
f`z=1+a(2z+3)=0
f`w=1+a(2w+4)=0
-1/(2x+1)=-1/(2y+2)=-1/(2z+3)=-1/(2w+4)
y=(2x-1)/2
z=x-1
w=(2x-3)/2
x2+y2+z2+w2+x+2y+3z+4w=17/2
x²+(2x-1)²/4+(x-1)²+(2x-3)²/4+x+2x-1+3(x-1)+2(2x-3)=17/2
解出x即可,应该是两个值