82问答网 > 先化简，再求值 [1⼀(a-2)+(a눀-1)⼀(a+2)(a-1)]÷[a⼀(a+2)]눀 其中a=2005

先化简，再求值 [1⼀(a-2)+(a눀-1)⼀(a+2)(a-1)]÷[a⼀(a+2)]눀 其中a=2005

2025-04-14 05:40:42

推荐回答（2个）

回答1：

原式={ 1/(a-2)+(a+1)(a-1)/[(a+2)(a-1)] }×[ (a+2)/a ]²
=[ 1/(a-2)+(a+1)/(a+2) ]×[ (a+2)²/a² ]
={ (a+2)/[(a+2)(a-2)]+(a+1)(a-2)/[(a+2)(a-2)] }×[ (a+2)²/a² ]
={ [(a+2)+(a+1)(a-2)]/[(a+2)(a-2)] }×[ (a+2)²/a² ]
={ (a+2+a²-a-2)/[(a+2)(a-2)] }×[ (a+2)²/a² ]
={ a²/[(a+2)(a-2)] }×[ (a+2)²/a² ]
=(a+2)/(a-2)
当a=2005时
原式=(2005+2)/(2005-2)
=2007/2003

回答2：

[1/(a-2)+(a²-1)/(a+2)(a-1)]÷[a/(a+2)]²
=[1/(a-2)+(a+1)/(a+2)]*(a+2)^2/a^2
=[a^2/(a-2)(a+2)]*(a+2)^2/a^2
=(a+2)/(a-2) a=2005
=2007/203