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> 已知a、b、c都是整数,且对一切实数x,(x-a)(x-2002)-2=(x-b)(x-c)都成立,则这样的有序数组(a,b,c)共有______组.-数学
已知a、b、c都是整数,且对一切实数x,(x-a)(x-2002)-2=(x-b)(x-c)都成立,则这样的有序数组(a,b,c)共有______组.-数学
题目简介
已知a、b、c都是整数,且对一切实数x,(x-a)(x-2002)-2=(x-b)(x-c)都成立,则这样的有序数组(a,b,c)共有______组.-数学
题目详情
已知a、b、c都是整数,且对一切实数x,(x-a)(x-2002)-2=(x-b)(x-c)都成立,则这样的有序数组(a,b,c)共有______组.
题型:填空题
难度:中档
来源:不详
答案
∵(x-a)(x-2002)-2=(x-b)(x-c),
∴x2-(a+2002)x+2002a-2=(x+b)(x+c),
∵对一切实数x,(x-a)(x-2002)-2=(x-b)(x-c)都成立,
∴b与c是方程x2-(a+2002)x+2002a-2=0①的两整数根,
∵a是整数,
∴△=(a+2002)2-4(2002a-2)=(a-2002)2+8是完全平方,
令(a-2002)2+8=n2,这里n为正整数,n>|a-2002|.
于是有(n+a-2002)(n-a+2002)=8,
n+a-2002=2
n-a+2002=4
或
n+a-2002=4
n-a+2002=2
,
解得n=3,a=2001或2003;
从而方程①的两根为:[(a+2002)±3].
当a=2001时,方程①的两根为2000,2003;
当a=2003时,方程①的两根为2001,2004.故
满足条件的有序组(a,b,c)共有如下4组:
(2001,2000,2003),(2001,2003,2000),(2003,2001,21304),(2003,2004.2001).
故答案为:4.
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题目简介
已知a、b、c都是整数,且对一切实数x,(x-a)(x-2002)-2=(x-b)(x-c)都成立,则这样的有序数组(a,b,c)共有______组.-数学
题目详情
答案
∴x2-(a+2002)x+2002a-2=(x+b)(x+c),
∵对一切实数x,(x-a)(x-2002)-2=(x-b)(x-c)都成立,
∴b与c是方程x2-(a+2002)x+2002a-2=0①的两整数根,
∵a是整数,
∴△=(a+2002)2-4(2002a-2)=(a-2002)2+8是完全平方,
令(a-2002)2+8=n2,这里n为正整数,n>|a-2002|.
于是有(n+a-2002)(n-a+2002)=8,
解得n=3,a=2001或2003;
从而方程①的两根为:[(a+2002)±3].
当a=2001时,方程①的两根为2000,2003;
当a=2003时,方程①的两根为2001,2004.故
满足条件的有序组(a,b,c)共有如下4组:
(2001,2000,2003),(2001,2003,2000),(2003,2001,21304),(2003,2004.2001).
故答案为:4.