几道中学数学题, ☆答对加10分一道☆

来源:百度知道 编辑:UC知道 时间:2024/06/27 11:01:19
要有过程!
(1+1/1997+1/1999+1/2001)*(1/1997+1/1999+1/2001+1/2003)-(1+1/1997+1/1999+1/2001+1/2003)*(1/1997+1/1999+1/2001)

已知|a|=-a,b<0 ,化简|2a+4b|/(a+2b)^2-4/|a+2b|-2/|4b+3-|2a-3||

试求|x-1|+|x+2|+|x-3|+..+|x-2003|的最小值

1、
(1+1/1997+1/1999+1/2001)*(1/1997+1/1999+1/2001+1/2003)-(1+1/1997+1/1999+1/2001+1/2003)*(1/1997+1/1999+1/2001)
用乘法分配率
=[(1/1997+1/1999+1/2001+1/2003)+(1/1997+1/1999+1/2001)*(1/1997+1/1999+1/2001+1/2003)]-[(1/1997+1/1999+1/2001)+(1/1997+1/1999+1/2001)*(1/1997+1/1999+1/2001+1/2003)]
=(1/1997+1/1999+1/2001+1/2003)+(1/1997+1/1999+1/2001)*(1/1997+1/1999+1/2001+1/2003)-(1/1997+1/1999+1/2001)-(1/1997+1/1999+1/2001)*(1/1997+1/1999+1/2001+1/2003)
=(1/1997+1/1999+1/2001+1/2003)-(1/1997+1/1999+1/2001)
=1/1997+1/1999+1/2001+1/2003-1/1997-1/1999-1/2001
=1/2003

2、因为|a|=-a, 所以a≤0,又b<0,
所以 |2a+4b|<0 ,|a+2b|<0,|2a-3|<0
|2a+4b|/(a+2b)^2-4/|a+2b|-2/|4b+3-|2a-3||
=-(2a+4b)/(a+2b)^2 + 4/(a+2b) -2/|4b+3+2a-3|
=-(2a+4b)/(a+2b)^2 + 4/(a+2b) + 2/2a+4b
=-(2a+4b)/(a+2b)^2 + 4(a+2b)/(a+2b)^2 + 2/2a+4b
=2(a+2b)/(a+2b)^2 + 2/2a+4b
=2/(a+2b) + 1/(a+2b)
=3/(a+2b)

3、由绝对值的几何意义有,|x-1|表示,点x到点1的距离!
那么|x