Question

已知xyz=1,x+y+z=2,x^2+y^2+z^2=16.求代数式1\(xy+2z)+1\(yz+2z)+1\(zx

Answer 1

楼主是不是打错一个字母,第二项数2z应该是2x吧?

xy+2z=xy+4-2x-2y=(x-2)(y-2).

yz+2x=(y-2)(z-2),zx+2y=(z-2)(x-2).

4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx),

xy+yz+zx=-6.

(x-2)(y-2)(z-2)=xyz-2(xy+yz+zx)+4(x+y+z)-8=13.

原式=[(x-2)+(y-2)+(z-2)]/(x-2)(y-2)(z-2)

=-4/13.