1. ## inequality

Find the solutions set of the inequality $|x-2|<\frac{1}{x}$ , where x is not 0 .

THanks .

2. it's quite obvious that $x\ne0,$ that's an assumption that it must be implicit and that you should know.

well then, by a well known property your inequality equals $-\frac{1}{x} so you have to solve each one.

i'll help ya with the first one:

\begin{aligned}
x-2&>-\frac{1}{x} \\
x-2+\frac{1}{x}&>0 \\
\frac{x^{2}-2x+1}{x}&>0,
\end{aligned}

the numerator is a perfect square, but it has a critical point at $x=1,$ but we don't actually want that, since that'd turn the numerator zero and the inequality won't be satisfied, so to solve the previous inequality we just require that $x>0,$ but since we've stated that $x\ne1,$ the first solution set is $(0,1)\cup(1,\infty).$

apply the same procedure for the second inequality, that'd yield a second solution set, which you need to intersect with the first one to get the full solution set.

3. If $x<0$, the inequality has no solution.

If $x>0$ the inequality is $x|x-2|<1$

1) $x\in(0,2)\Rightarrow x(2-x)<1\Rightarrow (x-1)^2>0$, which is true.

2) $x\in[2,\infty)\Rightarrow x^2-2x-1<0\Rightarrow x\in(1-\sqrt{2},1+\sqrt{2})\cap [2,\infty)\Rightarrow x\in[2,1+\sqrt{2})$

From 1) and 2), $x\in(0,2)\cup[2,1+\sqrt{2})\Rightarrow x\in(0,1+\sqrt{2})$