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
=-1,1,|a+1/2| +(b-3)^2=0
a=-1/2,b=3
〔(2a+b)^2+(2a+b)*(b-2a)-6b]/(2b)
=[(-1+3)^2+(-1+3)(3+1)-18]/(2*3)
=[4+8-18]/6=-1,2,|a+1/2| +(b-3)^2=0
则
a+1/2=0
a=-1/2
b-3=0
b=3
〔(2a+b)^2+(2a+b)*(b-2a)-6b]除2b
=[(2a+b)(2a+b+b-2a)-6b]/2b
=[(2a+b)*2b-6b]/2b
=2a+b-3
=2*(-1/2)+3-3
=-1,1,