已知x²-5x-1997=0,求代数式[(x-2)³-(x-1)²+1]/(x-2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 20:52:08
![已知x²-5x-1997=0,求代数式[(x-2)³-(x-1)²+1]/(x-2)](/uploads/image/z/1765380-12-0.jpg?t=%E5%B7%B2%E7%9F%A5x%26sup2%3B-5x-1997%3D0%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F%5B%EF%BC%88x-2%EF%BC%89%26sup3%3B-%EF%BC%88x-1%EF%BC%89%26sup2%3B%2B1%5D%2F%28x-2%29)
x){}K+ԊKuM+t
--m
tmlz{~BN2ck]076 *дI*ҧv6p壎A%@%lM+A0hnjBTӍb-\%x6B]Ԑ@RhsI nB0
OjTVٽ m
iƐ0
F 1 )
已知x²-5x-1997=0,求代数式[(x-2)³-(x-1)²+1]/(x-2)
已知x²-5x-1997=0,求代数式[(x-2)³-(x-1)²+1]/(x-2)
已知x²-5x-1997=0,求代数式[(x-2)³-(x-1)²+1]/(x-2)
∵x²-5x-1997=0
∴x²=5x+1997
∴[(x-2)³-(x-1)²+1]/(x-2)
=(x-2)²-[(x-1)²-1]/(x-2)
又(x-2)²=x²-4x+4=5x+1997-4x+4=x+2001
(x-1)²-1=x²-2x=5x+1997-2x=3x+1997
∴原式=x+2001-(3x+1997)/(x-2)
=2001+(x²-2x-3x-1997)/(x-2)
=2001+(x²-5x-1997)/(x-2)
=2001+0
=2001