Question

数学！！！？？

Answer 1

1.f(x)=|2x-7|+1<=|x-1|

(1)x>=7/2:
2x-7+1<=x-1
x<=5
即有7/2<=x<=5
(2)1=<x<7/2:
7-2x+1<=x-1
x>=3
即有3<=x<7/2
(3)x<1:
7-2x+1<=1-x
x>=7
无解.
综上所述,解是3<=x<=5.
2,存在X,有f(x)=|2x-7|+1<=ax
(1)x>=7/2:
2x-7+1<=ax
a>=2-6/x
故有a>=2-6/(7/2)=2-12/7=2/7,即有a>=2/7
(2)0<x<7/2:
7-2x+1<=ax
a>=8/x-2
故有a>=8/(7/2)-2=16/7-2=2/7
(3)x<0:
7-2x+1<=ax
a<=8/x-2
故有a<=-2
综上所述,范围是a>=2/7或a<=-2

Answer 2

f(x)<=|x-1|
|2x-7|-|x-1|+1<=0
当x<=1时 7-2x-1+x+1<=0
x>=7(舍去)
当1<x<=7/2时 7-2x-x+1+1<=0
x>=3
所以3<=x<=7/2
当x>7/2时 2x-7-x+1+1<=0
x<=5
所以7/2<x<=5
综上所述解集为{x|3<=x<=5}

Answer 3

f(x)<=|x-1|
|2x-7|-|x-1|+1<=0
当x<=1时 7-2x-1+x+1<=0
x>=7(舍去)
当1<x<=7/2时 7-2x-x+1+1<=0
x>=3
所以3<=x<=7/2
当x>7/2时 2x-7-x+1+1<=0
x<=5
所以7/2<x<=5
综上所述解集为{x|3<=x<=5}

Answer 4

1.(1)当2x-7>=0时,即x>=7/2 2x-7+1<=x-1 x<=5 7/2<=x<=5 (2)当2x-7<=0且x-1>=0时,即1<=x<=7/2 x>=3 3<=x<=7/2 (3)当x-1<=0时,即x<=1 x>=7 不符合 综上 3<=x<=5
2.a>=2/7

Answer 5

1、f(x)>=1
分俩种情况：x>=1时，
1-x<=f(x)<=x-1
2x>=7与2<=2x<7

x<1时，
f(x)>x-1,f(x)<1-x

Answer 6

1.3≤x≤5
2.-2≤a≤2/7