若(4a+3b)的绝对值+(3b+2)^2=0,求多项式2(2a+3b)^2-3(2a+3b)+8(2a+3b)^2-7

|4a+3b|+(3b+2)²=0
因为|4a+3b|≥0,(3b+2)²≥0
要使它们的和等于0,则两个数都要等于0
所以|4a+3b|=0,(3b+2)²=0
4a+3b=0,3b+2=0
解得 a=1/2,b=-2/3
所以 2a+3b=2×(1/2)+3×(-2/3)=1-2=-1
2(2a+3b)²-3(2a+3b)+8(2a+3b)²-7(2a+3b)
=10(2a+3b)²-10(2a+3b)
=10(2a+3b)(2a+3b-1)
=10×(-1)×[(-1)-1]
=20