先化简,再求值:3a^2b-[2ab^2-2(ab-3/2a^2b)]+2ab.其中a、b满足|a+3b+1|+(2a+4)^2=0

问题描述:

先化简,再求值:3a^2b-[2ab^2-2(ab-3/2a^2b)]+2ab.其中a、b满足|a+3b+1|+(2a+4)^2=0

因为|a+3b+1|+(2a+4)²=0,两个非负数的和为0,这两个非负数都为0
所以a+3b+1=0且2a+4=0
a=-2,b=1/3
原式=3a²b-(2ab²-2ab+3a²b)+2ab
=3a²b-2ab²+2ab-3a²b+2ab
=(3a²b-3a²b)-2ab²+(2ab+2ab)
=0-2ab²+4ab
=-2ab²+4ab
当a=-2,b=1/3时
原式=-2×(-2)×(1/3)²+4×(-2)×1/3
=-2×(-2)×1/9+4×(-2)×1/3
=4/9-8/3
=4/9-24/9
=-20/9