已知x,y,z为有理数A=2x³-xyz,B=y³-z²+xyz,C= -x³+2y²-xyz,且(x+1)²+(y-1)+|z|=0 求A-[2B-3(C-A)]
问题描述:
已知x,y,z为有理数A=2x³-xyz,B=y³-z²+xyz,C= -x³+2y²-xyz,且(x+1)²+(y-1)+|z|=0 求A-[2B-3(C-A)]
(x+1)²+(y-1)²+|z|=0
答
(x+1)²+(y-1)²+|z|=0
(x+1)²=0,(y-1)²=0,|z|=0
x=-1
y=1
z=0
A=2x³-xyz
=2*(-1)³-(-1)*1*0
=-2
B=y³-z²+xyz
=1³-0²+(-1)*1*0
=1
C= -x³+2y²-xyz
= -(-1)³+2*1²-(-1)*1*0
=1+2
=3
A-[2B-3(C-A)]
=A-[2B-3C+3A]
=A-2B+3C-3A
=-2A-2B+3C
=-2*(-2)-2*1+3*3
=4-2+9
=11