先化简[(a-1)分之1-(a+1)分之1]÷(2a²-2)分之a

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先化简[(a-1)分之1-(a+1)分之1]÷(2a²-2)分之a
先化简[(a-1)分之1-(a+1)分之1]÷(2a²-2)分之a

先化简[(a-1)分之1-(a+1)分之1]÷(2a²-2)分之a
原式=(a+1-a+1)/(a+1)(a-1)÷a/2(a+1)(a-1)
=2/(a+1)(a-1)×2(a+1)(a-1)/a
=4/a

[(a-1)分之1-(a+1)分之1]÷(2a²-2)分之a
=[(a²-1)分之(a+1)-(a²-1)分之(a-1)]÷(2a²-2)分之a
=[(a²-1)分之(a+1-a+1)]÷[2(a²-1)分之a]
=[(a²-1)分之2]×[a分之2(a²-1)]
=a分之4