已知(a-2)²+（b+1）²=0，求代数式3a²b+ab²-3a²b+5ab+ab²-4ab+1⼀2a²b

因为一个数的平方大于等于0
所以只有当 a - 2 = 0 且 b + 1 = 0 时等号才成立
所以 a = 2 , b = -1
所以 3a²b + ab² - 3a²b + 5ab + ab² - 4ab + 1/2a²b
= 2ab² + (1/2)a²b + ab
= ab(2b + a/2 + 1)
= 2×(-1)×[2×(-1) +1 + 1]
= (-2)×0
= 0

(a-2)²+（b+1）²=0
(a-2)²=0,（b+1）²=0
a=2,b=-1

3a²b+ab²-3a²b+5ab+ab²-4ab+1/2a²b
=3a²b-3a²b+1/2a²b+ab²+ab²+5ab-4ab
=1/2a²b+2ab²+ab
=ab(1/2a+2b+1)
=2*(-1)*[1/2*2+2*(-1)+1]
=-2*(1-2+1)
=-2*0
=0

(a-2)²+（b+1）²=0
a=2 b=-1

3a²b+ab²-3a²b+5ab+ab²-4ab+1/2a²b
=1/2a²b+2ab²+ab
=1/2(-4)+2(2)+(-2)
=-2+4-2
=0

由已知得a=2 b=-1
其他的会了吧？