Question

一道数学的分式计算题目，帮我，【a的平方-4a+4分之（a+1）（a+2）】除以a-2分之a，要过程，救命的啊

拜托了，而且一定要是正确的哦
是[a^2减去4a+4分之（a+1）（a+2）再减去a的平方-2a分之a] 再除以a-2分之a，照着我叙述的写

Answer 1

[(a+1)(a+2)/(a^2-4a+4)-a/(a^2-2a)]/[a/(a-2)]
=[(a+1)(a+2)/(a-2)^2-a/a(a-2)]/[a/(a-2)]
=[(a+1)(a+2)/(a-2)^2-1/(a-2)]*(a-2)/a
=(a+1)(a+2)/(a-2)^2*(a-2)/a-1/(a-2)*(a-2)/a
=(a+1)(a+2)/a(a-2) -1/a
=(a+1)(a+2)/a(a-2) -(a-2)/a(a-2)
=[(a+1)(a+2) -(a-2)]/a(a-2)
=(a^2+3a+2-a+2)/a/(a-2)
=(a^2+2a+4)/a(a-2)

Answer 2

{ [(a+1)(a+2) / (a^2 - 4a +4) ]- [ a / (a^2 - 2a) ] } / [ a / (a-2) ]
= { [(a+1)(a+2) / (a-2)^2 ] - [ a / a(a - 2) ] } * [ (a-2) / a ]
=[(a+1)(a+2) / a(a-2) ] - [ (a-2) / a(a - 2) ]
=( a^2 + 2a + 4 ) / a(a - 2)

Answer 3

[(a+1)(a+2)/(a^2-4a+4)]/[a/(a-2)]
=[(a+1)(a+2)/(a-2)^2]/[a/(a-2)]