Is the exact value of any likelihood meaningless? While reading about likelihood, I have heard that "the exact value of any likelihood is meaningless" why?
So, because of that we may use the likelihood ratio.
So, my question is, why the exact value of likelihood is meaningless? And what is the benefit of likelihood-ratio over the likelihood?

The likelihood function is usually taken to be the PDF viewed as
as a function of parameters for known data.
For example, if I have
a coin with Heads probability $\theta$ and toss it $n = 10$ times, getting $x = 3$ heads, then I can take the likelihood function to be
${n\choose x}\theta^x(1-\theta)^{n-x},$ considered as a function of $\theta.$
If I want the MLE $\hat \theta$ of $\theta,$ then I might write
the likelihood function as $$f(\theta \mid  x = 3)\propto \theta^3(1-\theta)^7,$$ where the symbol $\propto$ (read as "proportional to")
indicates that the constant ${n\choose x} = {10\choose 3} = 120$
is omitted. The maximum of the likelihood function is at
$\hat \theta = x/n = 0.3,$ whether I use or ignore the constant
${10\choose 3}.$
So the values of the likelihood function might be considered
less important than its shape, which leads to the MLE $\hat\theta.$

