若|x-1|+(y+3)²+√(根号)x+y+z=0,求xyz的值.
问题描述:
若|x-1|+(y+3)²+√(根号)x+y+z=0,求xyz的值.
答
∵|x-1|+(y+3)²+√(根号)x+y+z=0
∴x-1=0 y+3=0 x+y+z=0
x=1 y=-3 z=2
∴xyz=1*(-3)*2=-6
答
|x-1|+(y+3)²+√(根号)x+y+z=0
所以 x-1=0 x=1
y+3=0 y=-3
x+y+z=0 z=-(x+y)=2
xyz=1*-3*2
=-6