先化简,再求值:[3/(x-1)-x-1]/[(x-2)/(x²-2x+1)],其中,x=-根号2
展开全部
[3/(x-1)-x-1]/[(x-2)/(x²-2x+1)]
=[3/(x-1)-(x+1)(x-1)/(x-1)]/[(x-2)/(x-1)²]
=(3-x²+1)/(x-1)]/[(x-2)/(x-1)²]
=(4-x²)/(x-1)]/[(x-2)/(x-1)²]
=-(x²-4)/(x-1)]/[(x-2)/(x-1)²]
=-[(x-2)(x+2)/(x-1)]/[(x-2)/(x-1)²]
=-(x-2)(x+2)/(x-1)*(x-1)²/(x-2)
=-(x+2)(x-1)
=-(-√2+2)(-√2 -1)
=(-√2+2)(√2+1)
=-2-√2+2√2+2
=√2
=[3/(x-1)-(x+1)(x-1)/(x-1)]/[(x-2)/(x-1)²]
=(3-x²+1)/(x-1)]/[(x-2)/(x-1)²]
=(4-x²)/(x-1)]/[(x-2)/(x-1)²]
=-(x²-4)/(x-1)]/[(x-2)/(x-1)²]
=-[(x-2)(x+2)/(x-1)]/[(x-2)/(x-1)²]
=-(x-2)(x+2)/(x-1)*(x-1)²/(x-2)
=-(x+2)(x-1)
=-(-√2+2)(-√2 -1)
=(-√2+2)(√2+1)
=-2-√2+2√2+2
=√2
本回答由提问者推荐
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
