1.a-2分之a+1·a分之a-1-a^2-2a分之4a-5 2.x^2-x分之x^2+2x+1·x+1分之x-x-1分之1

(a+1)/(a-2)*(a-1)/a-(4a-5)/(a^2-2a)
=(a+1)(a-1)/[a(a-2)]-(4a-5)/[a(a-2)]
=(a^2-1)/[a(a-2)]-(4a-5)/[a(a-2)]
=[(a^2-1)-(4a-5)]/[a(a-2)]
=(a^2-1-4a+5)/[a(a-2)]
=(a^2-4a+4)/[a(a-2)]
=(a-2)^2/[a(a-2)]
=(a-2)/a
(x^2+2x+1)/(x^2-x)*x/(x+1)-1/(x-1)
=(x+1)^2/[x(x-1)]*x/(x+1)-1/(x-1)
=(x+1)/(x-1)-1/(x-1)
=(x+1-1)/(x-1)
=x/(x-1)
(a+3)/(a-2)-1/a÷(2-a)/(9-a^2)
=(a+3)/(a-2)-1/a÷(a-2)/(a^2-9)
=(a+3)/(a-2)-1/a*(a^2-9)/(a-2)
=a(a+3)/[a(a-2)]-(a^2-9)/[a(a-2)]
=[a(a+3)-(a^2-9)]/[a(a-2)]
=[a^2+3a-a^2+9]/[a(a-2)]
=(3a+9)/[a(a-2)]
=3(a+3)/[a(a-2)]
=3*(5/2+3)/[5/2*(5/2-2)]
=(3*11/2)/(5/2*1/2)
=(33/2)/(5/4)
=33/2*4/5
=66/5hey 还有一道题目， x-1 分之2 + 1- x分之（）=1 填空，我多加你10分！2/(x-1)+()/(1-x)=1 2/(x-1)-()/(x-1)=1 [2-()]/(x-1)=1 2-()=x-1 ()=2+1-x ()=3-x