Question

先化简代数式（a+2分之a、+、a-2分之2）除以a的平方-4分之1。然后取一个合适的a值，代入求值 急！！

Answer 1

[a/(a＋2)＋2/(a－2)]÷1/(a²－4)
＝(a²－2a＋2a＋4)/(a²－4)×(a²－4)
＝a²＋4
当a＝1时
原式＝1＋4＝5
a≠±2

Answer 2

解：[a/(a+2)+2/(a-2)]÷1/(a²-4)
=[a/(a+2)+2/(a-2)]×(a+2)(a-2)
=a/(a+2)×(a+2)(a-2)+2/(a-2))×(a+2)(a-2)
=a(a-2)+2(a+2)
=a²-2a+2a+4
=a²+4
解法二：[a/(a+2)+2/(a-2)]÷1/(a²-4)
=[a(a-2)/(a+2)(a-2)+2(a+2)/(a+2)(a-2)]×(a+2)(a-2)
=[(a²-2a+2a+4)/(a+2)(a-2)]×(a+2)(a-2)
=a²+4
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