Question

Write the solution set of $|x+\frac{1}{x}|>2$

Answer 1

$|x+\frac{1}{x}|>2\Rightarrow x+\frac{1}{x}<-2orx+\frac{1}{x}>2Case1:Whenx+\frac{1}{x}<-2x+\frac{1}{x}<-2\Rightarrow \frac{x^2+2x+1}{x}<0\Rightarrow \frac{(x+1{)}^2}{x}<0(x+1)positiveforx>-1,(x+1)negativeforx<0,xzeroforx=0\Rightarrow \frac{(x+1{)}^2}{x}<0,forx<-1or\frac{(x+1{)}^2}{x}<0for-1<x<0Case2:Whenx+\frac{1}{x}>2x+\frac{1}{x}>2\Rightarrow \frac{x^2-2x+1}{x}>0\Rightarrow \frac{(x-1{)}^2}{x}>0(x-1)positiveforx>1,(x-1)negativeforx<1,(x-1)=0forx=1xpositiveforx>0,xnegativeforx<0,xzeroforx=0\Rightarrow \frac{(x-1{)}^2}{x}>0for0<x<1or\frac{(x-1{)}^2}{x}>0forx>1\text{Combining 1 and 2,we obtain that the solution set of the given inequation is}x<-1or-1<x<0or0<x<1orx>1\Rightarrow xϵR=\{-1,0,1\}$