Question

已知x2-5-1999=0,求代数式（x-2）³-（x-1）²＋1除以x-2的值

Answer 1

x2-5-1999=0应该改为x2-5x-1999=0过程如下：((x-2)³-(x-1)²＋1)/(x-2)=(x-2)^2-((x-1)^2-1)/(x-2)=(x-2)^2-x(x-2)/(x-2)=x^2-4x+4-x=x^2-5x+4∵x^2-5x-1999=0∴x^2-5x=1999∴((x-2)³-(x-1)²＋1)...