1.[a/(ab-b²）-b/(a²-ab)]÷[1+(a²+b²)/2ab],其中a=-1+根号3,b=-1-根号3

题目

1.[a/(ab-b²）-b/(a²-ab)]÷[1+(a²+b²)/2ab],其中a=-1+根号3,b=-1-根号3
2.解关于x的方程：(x+1)/(a+b)+(x-1)/(a-b)=2a/(a²-b²) (a≠0)
3.已知A·B=2x+8,A=3x/(x-2)-x/(x+2),求B

答案

1.[a/(ab-b²）-b/(a²-ab)]÷[1+(a²+b²)/2ab],其中a=-1+根号3,b=-1-根号3=[a/b(a-b)-b/a(a-b)]÷(2ab+a²+b²)/2ab=(a²-b²)/ab(a-b)÷(a+b)²/2ab=2/(a+b)=2/(-1+√3-1-√...