先化简,再求值(a的平方+1/a+2)÷(a+2)(a-1)/a的平方+2a,其中a的平方-4=0
1个回答
展开全部
(a的平方+1/a+2)÷(a+2)(a-1)/a的平方+2a
=(a^2+1)/(a+2)*a(a+2)/(a+2)(a-1)
=[(a^2+1)/(a+2)]*[a/(a-1)]
a^2-4=0
a^2=4
a=2,a=-2
由题知,分母为a+2,不等于0,a不等于-2,
取a=2
原题:
=[(a^2+1)/(a+2)]*[a/(a-1)]
=[(2^2+1)/(2+2)]*2/(2-1)
=(5/4)*2
=5/2
=(a^2+1)/(a+2)*a(a+2)/(a+2)(a-1)
=[(a^2+1)/(a+2)]*[a/(a-1)]
a^2-4=0
a^2=4
a=2,a=-2
由题知,分母为a+2,不等于0,a不等于-2,
取a=2
原题:
=[(a^2+1)/(a+2)]*[a/(a-1)]
=[(2^2+1)/(2+2)]*2/(2-1)
=(5/4)*2
=5/2
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询