已知x²-5x-2007=0,则代数式((x-2)³-(x-1)²+1)⼀ x-2=?

2024-12-17 05:46:05
推荐回答(2个)
回答1:

解:∵x²-5x-2007=0
∴x²-5x=2007
[(x-2)³-(x-1)²+1]/x-2)
=[(x-2)³-x(x-2)]/(x-2)
=[(x-2)(x²-5x+4)]/(x-2)
=x²-5x+4
=2007+4
=2011.

回答2:

因为x²-5x-2007=0,
所以x²-5x=2007。
所以[(x-2)³-(x-1)²+1]/(x-2)
=[(x-2)³-(x²-2x+1)+1]/(x-2)
=[(x-2)³-x²+2x]/(x-2)
=(x-2)[(x-2)²-x]/(x-2)
=x²-4x+4-x
=x²-5x+4
=2007+4
=2011。