Electronic – rely on a the simulation of an op-amp based differential amplifier without looking at the op-amps common mode signal

My question is why engineers developed theoretically something called common-mode voltage And why they chose it particularly as (v1+v2)/2. What is the benefit of all these?

If you have two voltages, you could specify them any number of ways. The most obvious way would be simply:

$$ \begin{align} V_1 &= \text{something} \\ V_2 &= \text{something else} \end{align} $$

However, this isn't the most convienent method for all applications. Consider this circuit, which is a rather typical application of a differential amplifier:

^{simulate this circuit – Schematic created using CircuitLab}

A real circuit doesn't actually have capacitors C1 or C2, or inductors L2 or L3, but these are the unintentional capacitive and inductive couplings to other stuff (adjacent cables, that nearby computer monitor, distant RF radiators, ...) that your circuit must necessarily have by virtue of existing in a real environment.

Now, given the two voltages \$V_1\$ and \$V_2\$, we have the problem of figuring out what \$V_{signal}\$ was.

Well, since this is a differential amplifier, that's easy. It's the difference between the voltages, or the differential mode voltage:

$$ V_{dm} = V_2 - V_1 $$

But that's not enough information to know what the two voltages actually are. We need something else. That something else is the common mode voltage:

$$ V_{cm} = \frac{V1 + V2}{2} $$

You might wonder why this, when something simpler (such as any one of simply \$V_1\$ or \$V_2\$ would also do). The reason is that this is the "average" or "middle" or "center" voltage of \$V_1\$ and \$V_2\$, or in other words, the difference from \$V_1\$ or \$V_2\$ to \$V_{cm}\$ is the same:

$$ |V_1 - V_{cm}| = |V_2 - V_{cm}| $$

This has the convenient property that if \$V_1\$ and \$V_2\$ are switched, \$V_{cm}\$ remains the same.

This does what the OP wanted, a differential output around a defined output common mode, with no more, and in fact fewer, precision resistors.

^{simulate this circuit – Schematic created using CircuitLab}

If the common mode voltage does not match the input at Vcm, then OA3 drives an input voltage into both inverting inputs, with the same gain, which will cause both output voltages to move the same amount in the same direction, maintaining the existing differential gain, but shifting the common mode until there is no error.

Stability may be an issue, as there are two amps in a feedback loop. I suspect it would be easy to stabilise by clobbering the OA3 bandwidth, and/or speeding up OA1/2 a little with a small C across R3 and R5, which may or may not be desirable from the differential behaviour point of view.

Note that the only resistors that need to be matched are R1 and R2, which set the two output terminals to be equally disposed around Vcm. The differential gain is just (R3+R4+R5+R6)/(R4+R6), it does not need matched resistors, these can be four arbitrary value resistors, subject to getting the correct gain of course. I emphasise that fact by putting 4 unmatched values in the diagram for those resistors. The diff gain is 7 (21k/7k), with the outputs exactly disposed around Vcm because of R1==R2, and OA3. Try it!