已知x²-5x-2004=0,求代数式(x-2)³-(x-1)²+1/(x-2)的值
问题描述:
已知x²-5x-2004=0,求代数式(x-2)³-(x-1)²+1/(x-2)的值
答
x²-5x-2004=0x²-5x=2004(x-2)³-(x-1)²+1/(x-2)=(x³-6x²+12x-8-x²+2x-1+1)/(x-2)=(x³-7x²+14x-8)/(x-2)=x²-5x+4=2004+4=2008