Answered

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please help me                          llxl+2l=2

Sagot :

naǫ
[tex]||x|+2|=2[/tex]

if x≥0 then |x|=x
if x<0 then |x|=-x

1) for x≥0

[tex]|x+2|=2 \\ x+2=2 \ \lor \ x+2=-2 \\ x=0 \in D \ \lor \ x=-4 \notin D[/tex]


2) for x<0

[tex]|-x+2|=2 \\ -x+2=2 \ \lor \ -x+2=-2 \\ -x=0 \ \lor \ -x=-4 \\ x=0 \notin D \ \lor \ x=4 \notin D[/tex]

The answer is:
[tex]x=0[/tex]
There is only one solution to the problem and that is x=0.
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