x=2004,求|4x^2-4x+3|-4|x^2+2x+2|+13x+6的值
问题描述：
x=2004,求|4x^2-4x+3|-4|x^2+2x+2|+13x+6的值
x=2004|4x^2-4x+3|-4|x^2+2x+2|+13x+6
答
|4x^2-4x+3|-4|x^2+2x+2|+13x+6
两个绝对值符号内的式子△都小于0并且二次项系数都是大于0的
所以恒为正值
所以|4x^2-4x+3|-4|x^2+2x+2|+13x+6
=4x^2-4x+3-4x^2-8x-8+13x+6
=x+1
=2005