IX-1I+IY+ZI+IZ+3I=0,则(x+1)(y+5)(z-1)=?
问题描述:
IX-1I+IY+ZI+IZ+3I=0,则(x+1)(y+5)(z-1)=?
答
x-1=0 x=1 z+3=0 z=-3 y+z=0 y=3 (x+1)(y+5)(z-1)=2*8*(-4)=-64
IX-1I+IY+ZI+IZ+3I=0,则(x+1)(y+5)(z-1)=?
x-1=0 x=1 z+3=0 z=-3 y+z=0 y=3 (x+1)(y+5)(z-1)=2*8*(-4)=-64