先化简再求值：[1/(a-b)+1/(b-a)]除以ab/(a+b),其中a=(根号下2)+1,b=根号下2

1)原式=0.
2)原式=1/(x+1)×(x-1)(x+2)/[(x-1)(x+1)]+(x+2)/(x+1)
=(x+2)/(x+1)^2+(x+2)/(x+1)
=(x+2)(1+x+1)/(x+1)^2
=(x+2)^2/(x+1)^2.
3)x^2-x-1=0,
∴x=(1土√5)/2,x^2=x+1,
原式=[(x^2-1)/x+2/x]×x/(2x-1)+2x+1
=(x^2+1)/(2x-1)+2x+1
=5x^2/(2x-1)
=5(x+1)/(2x-1)
=5[(1土√5)/2+1]/(土√5)
=土√5[3土√5]/2
=(土3√5+5)/2.