Question

已知x^2-5x-2=0,则代数式(x-2)^3-(x-1)^2+1/x-2的值为

要快哦 必须要有过程哦。

Answer 1

解：
原所求代数式为：
[(x-2)³-(x-1)²+1]/(x-2)
=(x-2)²-[(x-1)²-1]/(x-2)
=(x-2)²-[(x-1+1)(x-1-1)]/(x-2)
=(x-2)²-x
=x²-4x+4-x
=x²-5x+4
由x²-5x-2=0可得：
x²-5x=2
因此：
[(x-2)³-(x-1)²+1]/(x-2)
=(x-2)²-[(x-1)²-1]/(x-2)
=(x-2)²-[(x-1+1)(x-1-1)]/(x-2)
=(x-2)²-x
=x²-4x+4-x
=x²-5x+4
=6

Answer 2

【（x-2）^3-（x-1）^2+1】/（x-2）
=(x-2)^2-(x^2-2x+1-1)/(x-2)
=x^2-4x+4-(x^2-2x)/(x-2)
=x^2-4x+4-x
=x^2-5x+4
=2+4
=6