已知x2-5-1999=0,求代数式（x-2）³-（x-1）²＋1除以x-2的值
2个回答
展开全部
x-2)(x-2)^2-(x-1)^2+1/(x-2)
=(x-2)(x-2)^2-[(x-1)^2-1]/(x-2)
=(x-2)(x-2)^2-[(x-1+1)*(x-1-1)]/(x-2)
=(x-2)(x-2)^2-[x*(x-2)]/(x-2)
=(x-2)[(x-2)^2-x]/(x-2)
=(x-2)^2-x
=x^2-5x+4
因为x^2-5x-1999=0
所以x^2-5x=1999
所以x^2-5x+4=1999+4=2003
=(x-2)(x-2)^2-[(x-1)^2-1]/(x-2)
=(x-2)(x-2)^2-[(x-1+1)*(x-1-1)]/(x-2)
=(x-2)(x-2)^2-[x*(x-2)]/(x-2)
=(x-2)[(x-2)^2-x]/(x-2)
=(x-2)^2-x
=x^2-5x+4
因为x^2-5x-1999=0
所以x^2-5x=1999
所以x^2-5x+4=1999+4=2003
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？
展开全部
x²-5x-1999=0
x²-5x=1999
[(x-2)³-(x-1)²+1]/(x-2)
=(x-2)² -[(x-1)²-1]/(x-2)
=(x-2)²-(x²-2x)/(x-2)
=(x-2)²-x(x-2)/(x-2)
=(x-2)²-x
=x²-4x+4-x
=x²-5x+4
=1999+4
=2003
x²-5x=1999
[(x-2)³-(x-1)²+1]/(x-2)
=(x-2)² -[(x-1)²-1]/(x-2)
=(x-2)²-(x²-2x)/(x-2)
=(x-2)²-x(x-2)/(x-2)
=(x-2)²-x
=x²-4x+4-x
=x²-5x+4
=1999+4
=2003
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
