Question

（a/ab-b²-b/a²-ab）/2ab+a²+b²/2ab,其中a=-1+根号三,b=-1-根号三

Answer 1

原式化为：通分（a/ab-b²-b/a²-ab）=>(a²-b²)/2a²b²(a-b) =>

（a+b）/(2a²b²)+a²+b²/(2ab)

a²=4-2根号三,b²=4+2根号三,a+b=-2,ab=-2

即 （-2）/[2(-2)(-2)]+4-2根号三+4+2根号三/(-4)=-1/4+4-2根号三-1-根号三/2

=11/4-5根号三/2