答案没错 x^1/2-y^1/2+z^1/2=0 把z移到右边然后square both sides x+y-2(xy)^1/2 = z x+y-z=2(xy)^1/2 继续square both sides x^2+y^2+z^2+2(xy-yz-xz)=4xy x^2+y^2+z^2=4xy-2(xy-yz-xz) =2(xy+yz+xz) -- by 会员 魏小妞要出国 (2010/8/15 22:44:31)
z^1/2=y^1/2-x^1/2 => z=y+x-2*(xy)^1/2 => 2(xy)=y+x-z y^1/2=x^1/2+z^1/2=> y=x+z+2*(xz)^1/2 => 2(xz)=y-x-z x^1/2=y^1/2-z^1/2 => x=y+z-2*(yz)^1/2 => 2(yz)=y-x+z 4xy = (y+x-z)^2 = x^2+y^2+z^2 +2xy-2xz-2yz 4xz = (y-x-z)^2 = x^2+y^2+z^2 -2xy+2xz-2yz 4xy = (y-x+z)^2 = x^2+y^2+z^2 -2xy-2xz+2yz 4xy+4xz+4yz = 3x^2+3y^2+3z^2-2xy-2xz-2yz 6xy+6xz+6yz = 3x^2+3y^2+3z^2 x^2+y^2+z^2 = 2xy+2xz+2yz 上面我算错了。 |