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

分类: 作业答案 • 2022-07-08 08:38:38

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

2005

答

原式=((2x-1)^2+2)-4((x+1)^2+1)+13x+6
=4x^2-4x+3-4x^2-8x-8+13x+6
=x+1
=2005

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