已知x^2-5x-1997=0,求代数式[（x-2)(x-2)-(x-1)(x-1)+1]/(x-2)的值...
问题描述：
已知x^2-5x-1997=0,求代数式[（x-2)(x-2)-(x-1)(x-1)+1]/(x-2)的值...
答
原式＝[x²－4－﹙x²－2x＋1﹚＋1]/﹙x－2﹚
＝﹙x²－4x＋4－x²＋2x－1＋1﹚/﹙x－2﹚
＝﹙﹣2x＋4﹚/﹙x－2﹚
＝﹣2﹙x－2﹚/﹙x－2﹚
＝﹣2
答
x^2-4x=x+1997,x^2-2x=3x+1997 所以：（x-2)(x-2)-(x-1)(x-1)+1]/(x-2) =(x+1997+4-3x-1997)/(x-2)=-2
答
[（x-2)(x-2)-(x-1)(x-1)+1]/(x-2)
={[(x-2)+(x-1)][(x-2)-(x-1)]+1}/(x-2)
=[-(2x-3)+1]/(x-2)
=(4-2x)/(x-2)
=-2
答
[（x-2)(x-2)-(x-1)(x-1)+1]/(x-2)=(4-2x)/(x-2)=-2