实数x、y、z满足x=6-3yx+3y-2xy+2z2=0,则x2y+z的值为_.

问题描述:

实数x、y、z满足

x=6-3y
x+3y-2xy+2z2=0
,则x2y+z的值为______.

x=6-3y                 ①
x+3y-2xy+2z2=0   ②

把①代入②中,可得:
6(y-1)2+2z2=0,
即y=1,z=0,
故x=3,
所以x2y+z=32=9,
故答案为9.