Question

已知XYZ=1,X+Y+Z=2,X的平方+Y的平方+Z的平方=16.求（XY+2Z)分之1+（YZ+2X)分之1+(ZX+2Y)分之1的值

Answer 1

No1. 将表达式分母化简，xy+2z=xy+4-2x-2y=(x-2)(y-2), 同理yz+2x=(y-2)(z-2) , zx+2y=(z-2)(x-2).
No2. 为通分准备 4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx), 则有 xy+yz+zx=-6 (x-2)(y-2)(z-2)=xyz-2(xy+yz+zx)+4(x+y+z)-8=13
No3. 通分，则有原式=[(x-2)+(y-2)+(z-2)] / [(x-2)(y-2)(z-2)] =-4/13

Answer 2

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.