Electronic – inside a General Purpose Operational Amplifier

Can you please explain what "mismatches in the differential input stage" is?

All op-amps have an input voltage error term. It's just like a small millivolt battery being placed in series with one of the inputs. If you build an amplifier with a gain of ten, the output voltage from the op-amp becomes offset from where you would expect it to be by 10 x the mV offset I just mentioned.

It is specifically this static error that the offset-voltage nulling circuit intends to eradicate. But, also remember that the offset voltage that is trying to be nulled, will drift with temperature (and time) and these drifting effects cannot be "cancelled" without re-positioning the potentiometer in your circuit.

This is easier than it looks. I won't solve it all the way, but lets call the voltage at the top of the pot \$V_a\$ and the voltage at the bottom of the pot, \$V_b\$. Lets call the positive input \$V_{1+}\$ and the negative, \$V_{1+}\$. The input terminals of the op amp are \$V_-\$ and \$V_+\$.

Using the rule that no current enters the op amp, it is straightforward to show that \$V_a=2V_- - V_{1+}\$. Using the rule the \$V_- = V_+\$, it is straightforward to show that \$V_b = 2V_- - V_{1-}\$. Now, we know the voltage at either side of the pot, and the value of the pot, so \$i_{pot} = \frac{(V_a-V_b)}{(1-X)R} = \frac{V_{1-} - V_{1+}}{(1-X)R}\$

Application of Kirchoff's current law to the node at the top of the pot should give you \$V_2\$