Find all possible values of x in #-1/(|x|-2)>=1#?

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Answer 1 · 2018-06-21T15:18:05

#1<=x<2# or #1<=x<2#

Multiplying by #-1# we have

#1/(|x|-2)<=-1#
a) let # x>=0# and #x>2# then we get

#x<=1# and we get #2x<x<=1# which is impossible.

b) #x<2# then we get #x>=1# and finally

#1<=x<2#

Next case:

#x<0#

then we get if #-2>x# #x>=-5/2#

and the interval #-5/2<=x<-2#

the last case is impossible.

Answer 2 · 2018-06-21T16:25:01

The solution is #x in (-2,-1]uu[1,2)#

The inequality is

#-1/(|x|-2)>=1#

Multiply both sides by #-1#

Therefore,

#1/(|x|-2)<=-1#

#<=>#, #1/(|x|-2)+1<=0#

So,

#{(1/(x-2)+1<=0),(1/(-x-2)+1<=0):}#

#<=>#, #{((1+x-2)/(x-2)<=0),((1-x-2)/(-x-2)<=0):}#

#<=>#, #{((x-1)/(x-2)<=0),((x+1)/(x+2)<=0):}#

Solving the #2# equation with a sign chart

#S={(x in[1,2)),(x in (-2,-1]):}#

Therefore,

The solution is #x in (-2,-1]uu[1,2)#

graph{1/(|x|-2)+1 [-10, 10, -5, 5]}