X
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XXX(x+1)(y+1)(z+1)=(xy+x+y+1)(z+1)=xyz+xz+yz+z+xy+x+y+1因为xyz+xz+yz+z+xy+x+y=230,所以:(x+1)(y+1)(z+1)=2
X
X
X
(x+1)(y+1)(z+1)
=(xy+x+y+1)(z+1)
=xyz+xz+yz+z+xy+x+y+1
因为xyz+xz+yz+z+xy+x+y=230,所以:
(x+1)(y+1)(z+1)=230+1=231
而231=11*7*3,且X
X=1,Y=4,Z=22 原式等价于(X+1)(Y+1)(Z+1)=230即三个自然数相乘,易知2*5*23=230所以得出XYZ分别为1,4,22