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If Av=R2/1R1 and open loop gain, G = 1e6 and open loop output impedance, Ro of say 300 Ohms, the Zout will become Ro/G *Av at DC
Since G changes with frequency as GBW is constant, Zout will also rise with f.
Effectively if any Vo occurs it is because the device does not have enough current drive to oppose the applied current. When it saturates externally, THe gain is high enough to force Vin(-) to match Vin(+) until it current limits or overheats.
In short: Both statements (Vx=0 and Rout=0) are not correct. It is only a simplification - and this fact should be mentioned.
In the shown circuit the opamp has negative resistive feedback with the consequences as follows:
* The voltage Vx at the inverting input is very small (Vout/Aol) because the open-loop gain Aol of the opamp is very large (1E4...1E6).
Therefore, we introduce just a very small error by setting Vx=0 during calculations (µVolts in reality).
* Something similar happens with the output resistance. Without feedback the output resistance, normally, is app. 50..200 Ohms.
However, due to negative feedback this resistance will be drastically reduced (divided by the loop gain) to values which mostly are below 1 Ohm.
For this reason and taking other uncertainties into account (tolerances) it is common practice to assume Rout=0.
Recommendation: Read some papers/contribution about feedback theory. Negative feedback is the most important technique for amplification.