Question

已知a+1/2的绝对值+b^2=6b-9求代数式[(2a+b)^2+(2a+b)(b-2a)-6ab]÷(2b)

Answer 1

|a+1/2|+b^2-6b+9=0 |a+1/2|+(b-3)^2=0

因为|a+1/2|大于等于0,(b-3)^2大于等于0

所以a+1/2=0,b-3=0

所以a=-1/2,b=3

原式=[4a^2+4ab+b^2-4a^2+b^2-6ab]÷2b

=[2b^2-2ab]÷2b

=b-a

=3-(-1/2)

=3又1/2