Question

b=根号[(2a-b)/(4a+3b)]+根号[(b-2a)/(4a+3b)]+3/2 ,则a²-ab+b&s

Answer 1

27/16

因为：根号[(2a-b)/(4a+3b)]、根号[(b-2a)/(4a+3b)]都成立

根号下必须>=0

所以：2a-b=0

b=3/2 => a=3/4

a²-ab+b²=27/16