Question

已知A=a³-3a²+2a-1,B=2a³+2a²-4a-5。当a=-1时，求3A-2[2B+(A-B)/2]的值。

已知A=a³-3a²+2a-1,B=2a³+2a²-4a-5。当a=-1时，求3A-2[2B+(A-B)/2]的值。

Answer 1

3A-2[2B+(A-B)/2]=3A-4B-A+B

=2A-3B
=2(a³-3a²+2a-1)-3(2a³+2a²-4a-5)
=(2a³-6a²+4a-2)-(6a³+6a²-12a-15)
=-4a³-12a²+16a+13
=4-12-16+13
=-11

Answer 2

A=a³-3a²+2a-1=-1-3-2-1=-7
B=2a³+2a²-4a-5=-2+2+4-5=-1

3A-2[2B+(A-B)/2]

=3×(-7)-2[2×(-1)+(-7+1)/2]
=-21-2(-2-3)
=-21+10
=-11

Answer 3

把a=-1带进去，得出A=-7，B=-1.再带近3A-2[2B+(A-B)/2]
算下就OK了。