Solving $\frac{\sqrt{r+1}-\sqrt{r-1}}{\sqrt{r+1}+\sqrt{r-1}}=\log_2\left(|x-2|+|x+2|\right)-\frac{11}{9}$, where $r=\frac{1+x^2}{2x}$

Hint:

Since $$\sqrt{\frac{1+x^2}{2x}\pm1}= \sqrt{(x\pm1)^2\over 2x} = {|x\pm1|\over \sqrt{2x}}$$ we have $$\frac{\sqrt{\frac{1+x^2}{2x}+1}-\sqrt{\frac{1+x^2}{2x}-1}}{\sqrt{\frac{1+x^2}{2x}+1}+\sqrt{\frac{1+x^2}{2x}-1}}={|x+1|-|x-1|\over |x+1|+|x-1|}$$

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Let $f(x)=|x+2|+|x-2| $, then $f(x)=4$ for $x\in[-2,2]$ and for $|x|>2$ we have $f(x)>4$.

So, for $|x|>2$ we have $\log_2f(x)-11/9>7/9$ so we have $|x+1|>8|x-1|$

a) if $x>2$ we get $x<9/7$ a contradiction and

b) if $x<-2$ we get $x>7/9$ a contradiction again.

So we are left with the $|x|\leq 2$ so we have $$|x+1|=8|x-1|$$

Can you finish?

Hint: multiplying the left-hand side's numerator and denominator by the numerator obtains $r-\sqrt{r^2-1}$, and this is just $\exp -|\ln x|$. So we want to solve $\exp -|\ln x|+\frac{11}{9}=\log_2 (|x-2|+|x+2|)$.

Simplify the LHS: $$\frac{\sqrt{\frac{1+x^2}{2x}+1}-\sqrt{\frac{1+x^2}{2x}-1}}{\sqrt{\frac{1+x^2}{2x}+1}+\sqrt{\frac{1+x^2}{2x}-1}}=\frac{\sqrt{\frac{(x+1)^2}{2x}}-\sqrt{\frac{(x-1)^2}{2x}}}{\sqrt{\frac{(x+1)^2}{2x}}+\sqrt{\frac{(x-1)^2}{2x}}}\overbrace{=}^{x>0}{|x+1|-|x-1|\over |x+1|+|x-1|}.$$ Hence: $${|x+1|-|x-1|\over |x+1|+|x-1|}=\log_2(|x-2|+|x+2|)-\frac{11}{9}.$$ Consider the intervals: $$x\in (0,1]: \qquad {2x\over 2}=\log_2 4-\frac{11}{9} \Rightarrow x=\frac79;\\ x\in (1,2]: \qquad {2\over 2x}=\log_2 4-\frac{11}{9} \Rightarrow x=\frac97;\\ x\in (2,+\infty): \qquad {2\over 2x}=\log_2 (2x)-\frac{11}{9} \Rightarrow \log_2x=\frac 1x+\frac29 \Rightarrow \emptyset, \text{because}:\\ \log_2x\ge 1>\frac{13}{18}\ge \frac 1x+\frac29, x\ge 2.$$