已知Ia+1I+(a+b-2)²=0,化简ab-2{ab-[3a²b-(4ab²+0.5a
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已知Ia+1I+(a+b-2)²=0,化简ab-2{ab-[3a²b-(4ab²+0.5ab)-3a²b]}3ab²,并求出它的值,
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
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
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