Question

．．．已知|a+1/2| +(b-3)^2=0,求代数式〔(2a+b)^2+(2a+b)*(b-2a)-6b]除2b的值

Answer 1

|a+1/2|+(b-3)²=0

因为绝对值与平方数都不小于0

所以有|a+1/2|=0,(b-3)²=0

a+1/2=0,b-3=0

解得a=-1/2,b=3

2a+b=2×(-1/2)+3=2

b-2a=3-2×(-1/2)=4

6b=18

2b=6

((2a+b)^2+(2a+b)*(b-2a)-6b)÷2b

=(2²+2×4-18)÷6

=(-6)÷6

=-1