The set of all values of x satisfying the inequality $$\log_{\left(x+\frac{1}{x}\right)}\left[\log_2\left(\frac{x-1}{x+2}\right)\right]>0$$ is

Given, $$\log_{\left(x+\dfrac{1}{x}\right)}\left[\log_2\left(\dfrac{x-1}{x+2}\right)\right]>0$$

The first constraint for this inequality is:

i.)$$x+\dfrac{1}{x}>0$$

or, $$\dfrac{x^2+1}{x}>0$$

Now, $$x^2+1$$ is always positive

So, for  $$\dfrac{x^2+1}{x}$$ to be positive, $$x$$ has to be positive i.e. $$x>0$$ ------>(1)

ii.)$$\log_2\left(\dfrac{x-1}{x+2}\right)>1$$

or, $$\dfrac{x-1}{x+2}>2^1$$

or, $$\dfrac{x-1}{x+2}-2>0$$

or, $$\dfrac{x-1-2x-4}{x+2}>0$$

or, $$\dfrac{-x-5}{x+2}>0$$

or, $$\left(x+5\right)\left(x+2\right)<0$$

or, $$-5<x<-2$$ ----->(2)

(1) and (2) are contradictory to each other

So, there is no value of $$x$$ possible satisfying this inequality

So, solution set is null set