已知X^2-5X-1991=0 则[(X-2)^4+(X-1)^2-1] / (X-1)(X-2) 的值为_______

来源:百度知道 编辑:UC知道 时间:2024/05/02 10:41:39
要求有过程,!

解:〔(x-2)4+(x-1)2-1〕=x4-8x3+25x2-34x+16

(x-1)(x-2)=x2-3x+2 两式相除=x2-5x+8 因为x2-5x-2005=0 x2-5x=2005

所以〔(x-2)4+(x-1)2-1〕/(x-1)(x-2)的值=2013

做错了,拿计数器按下最后1999

∵X^2-5X-1991=0
∴(x-2)²=x+1995 ,X^2=5x+1991
∴[(X-2)^4+(X-1)^2-1]/(x-2)(X-1)
=[(X-2)^4+x(x-2)]/(x-2)(X-1)
=(x-2)^3+x/(X-1)
=(x-2)(x+1995)+x/(X-1)

=x^2+1994x-3990/(X-1)

=5x+1991+1994x-3990/(X-1)

=1999x-1999/(X-1)
=1999