Răspuns :
x∈N*
x≠1
si
|2x-1|<5
x∈N*
x≠1
si
-5<2x-1<5
x∈N*
x≠1
si
-4<2x<6
x∈N*
x≠1
si
-2<x<3
x∈((-2;3)\{1} )∩N*={2}
x=2
x≠1
si
|2x-1|<5
x∈N*
x≠1
si
-5<2x-1<5
x∈N*
x≠1
si
-4<2x<6
x∈N*
x≠1
si
-2<x<3
x∈((-2;3)\{1} )∩N*={2}
x=2
[tex]|x-1|(|2x-1|-5)\ \textless \ 0,x\in \mathbb{N^*}\\
\text{Evident }|x-1|>0\text{pentru }x\neq 1,\text{ceea ce implica faptul ca }\\
|2x-1|-5\ \textless \ 0\\
|2x-1|\ \textless \ 5\\
-5\ \textless \ 2x-1\ \textless \ 5|+1\\
-4\ \textless \ 2x\ \textless \ 6|:2\\
-2\ \textless \ x\ \textless \ 3\\
x\in \mathbb{N^*}\Rightarrow x=2[/tex]