(a-2/a^2+2a-a-1/a^2-4a+4)÷a-4/a+2,其中a=√2-1 （求过程）
1个回答
展开全部
(a-2/a^2+2a-a-1/a^2+4a+4)÷a-4/a+2
=[(a-2)/a(a+2)-(a-1)/(a+2)^2] ×(a+2)/(a-4)
=[(a-2)/a-(a-1)/(a+2)] ×1/(a-4)
=(a^2-4-a^2+a)/a(a+2) ×1/(a-4)
=(a-4)/a(a+2) ×1/(a-4)
=1/a(a+2)
=1/(√2-1)(√2+1)
=1/(2-1)
=1
=[(a-2)/a(a+2)-(a-1)/(a+2)^2] ×(a+2)/(a-4)
=[(a-2)/a-(a-1)/(a+2)] ×1/(a-4)
=(a^2-4-a^2+a)/a(a+2) ×1/(a-4)
=(a-4)/a(a+2) ×1/(a-4)
=1/a(a+2)
=1/(√2-1)(√2+1)
=1/(2-1)
=1
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
