1. ## inequality

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

THanks .

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

well then, by a well known property your inequality equals $\displaystyle -\frac{1}{x}<x-2<\frac{1}{x},$ so you have to solve each one.

i'll help ya with the first one:

\displaystyle \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 $\displaystyle x=1,$ 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 $\displaystyle x>0,$ but since we've stated that $\displaystyle x\ne1,$ the first solution set is $\displaystyle (0,1)\cup(1,\infty).$

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 $\displaystyle x<0$, the inequality has no solution.

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

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

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

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