已知Ia+1I+(a+b-2)²=0,则有：
a+1=0且a+b-2=0
解得：a=-1,b=2-a=3
所以：
ab-2｛ab-[3a²b-(4ab²+0.5ab)-3a²b]｝+3ab²
=ab-2[ab-(3a²b-4ab²-0.5ab-3a²b)] +3ab²
=ab-2[ab-(-4ab²-0.5ab)] +3ab²
=ab-2(ab+4ab²+0.5ab) +3ab²
=ab-2(1.5ab+4ab²) +3ab²
=ab-3ab-8ab² +3ab²
=-2ab-5ab²
=-ab(2+5b)
=-(-1)*3*(2+15)
=3*17
=51