x*y*z=1 x+y+z=2 x^2+y^2+z^2=3 1/(xy+z-1)+1/(xz+y-1)+1/(yz+x-1)=?
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x*y*z=1 x+y+z=2 x^2+y^2+z^2=3 1/(xy+z-1)+1/(xz+y-1)+1/(yz+x-1)=?
x*y*z=1 x+y+z=2 x^2+y^2+z^2=3 1/(xy+z-1)+1/(xz+y-1)+1/(yz+x-1)=?
x*y*z=1 x+y+z=2 x^2+y^2+z^2=3 1/(xy+z-1)+1/(xz+y-1)+1/(yz+x-1)=?
1/(xy+z-1)=1/(xy+2-x-y-1)=1/[xy-x-(y-1)]=1/(x-1)(y-1)
同理1/(xz+y-1)=1/(x-1)(z-1), 1/(yz+x-1)=1/(y-1)(z-1)
1/(x-1)(y-1)+1/(x-1)(z-1)=[(z-1)+(y-1)]/(x-1)(y-1)(z-1)
=(y+z-2)/(x-1)(y-1)(z-1)=-x/(x-1)(y-1)(z-1)
-x/(x-1)(y-1)(z-1)+1/(y-1)(z-1)=(-x+x-1)/(x-1)(y-1)(z-1)=-1/(x-1)(y-1)(z-1)
(x-1)(y-1)(z-1)=xyz-xz-zy+z-xy+y+z-1=1+2-1-(xz+zy+xy)
(xz+zy+xy)=[(x+y+z)^2-(x^2+y^2+z^2)]/2=1/2
(x-1)(y-1)(z-1)=3/2
1/(xy+z-1)+1/(xz+y-1)+1/(yz+x-1)=-1/(x-1)(y-1)(z-1)=-2/3
用行列式的性质证明:y+z z+x x+y x y z x+y y+z z+x =2 z x y z+x x+y y+z y z x 这个怎么证?
(x-2y+z)(x+y-2z)分之(y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y) + (y+z-2z)(x-2y+z)分之(x-z)(y-z)=?第三部分那个是 (y+z-2x)(x-2y+z)分之(x-z)(y-z)
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