已知（a+2)（根号10+1）=1,求1+1/a-1-（1+2a）/（a²+a-2）的值

∵a+2=1/(√10+1)=(√10-1)/9
∴原式=1+1/（a-1)-(1+2a)/(a+2)(a-1)
=(a²+a-2+a+2-1-2a)/(a+2)(a-1)
=(a²-1)/(a+2)(a-1)
=(a-1)/(a+2)
=1-3/(a+2)
=1-(√10-1)/9
=(9-√10+1)/9
=(10-√10)/9