Where do these three dc error input voltage comes from?
Assume in the following circuit that \$V_-\$ and \$V_+\$ are at the exact same voltage, so that the differential is exactly zero volts:



simulate this circuit &ndash; Schematic created using CircuitLab
Then it follows that the collector currents are identical.
But it does not follow that the base currents are exactly the same, since the BJTs are different BJTs and they also may be at different temperatures, too. So, while we can for these purposes assume \$I_{\text{C}_1}=I_{\text{C}_2}\$, we must accept that \$I_{\text{B}_1}\ne I_{\text{B}_2}\$ since we have to assume that \$\beta_1\ne \beta_2\$.
There's another effect here. Even assuming that \$I_{\text{C}_1}=I_{\text{C}_2}\$ and recognizing that \$I_{\text{B}_1}\ne I_{\text{B}_2}\$, it's also likely that \$V_{\text{BE}_1}\ne V_{\text{BE}_2}\$. This is because the saturation currents of the two BJTs may also be different. So, while \$I_{\text{C}_1}=I_{\text{C}_2}\$ (our starting assumption that is required in order to ensure that \$V_-=V_+\$), we may have both \$I_{\text{B}_1}\ne I_{\text{B}_2}\$ and also \$V_{\text{BE}_1}\ne V_{\text{BE}_2}\$. Now, from the circuit we know absolutely that \$V_{\text{E}_1}= V_{\text{E}_2}\$, so the implication is that \$V_{\text{B}_1}\ne V_{\text{B}_2}\$.
If you are still following along, then I think you can see that:
$$\begin{align*}
V_1&amp;=V_{\text{B}_1}+I_{\text{B}_1}\cdot R_1\\\\
V_2&amp;=V_{\text{B}_2}+I_{\text{B}_2}\cdot R_2
\end{align*}$$
Please keep in mind here that the above equations are set up with the assumption that we've made all necessary adjustments to ensure that \$I_{\text{C}_1}=I_{\text{C}_2}\$ so that also \$V_-=V_+\$!
All we need to do now is to compute the difference between them:
$$\begin{align*}
V_1-V_2 &amp;= V_{\text{B}_1}+I_{\text{B}_1}\cdot R_1 - \left(V_{\text{B}_2}+I_{\text{B}_2}\cdot R_2\right)\\\\
&amp;=\left(V_{\text{B}_1} - V_{\text{B}_2}\right) + I_{\text{B}_1}\cdot R_{\text{B}_1}-I_{\text{B}_2}\cdot R_{\text{B}_2}\\\\
&amp;=\left(V_{\text{B}_1} - V_{\text{B}_2}\right) + \frac12\,I_{\text{B}_1}\cdot R_{\text{B}_1} + \frac12\,I_{\text{B}_1}\cdot R_{\text{B}_1} - \frac12\,I_{\text{B}_2}\cdot R_{\text{B}_2} - \frac12\,I_{\text{B}_2}\cdot R_{\text{B}_2}\\&amp;\hphantom{=\text{ } \left(V_{\text{B}_1} - V_{\text{B}_2}\right)}+\frac12\,I_{\text{B}_1}\cdot R_{\text{B}_2}- \frac12\,I_{\text{B}_1}\cdot R_{\text{B}_2} + \frac12\,I_{\text{B}_2}\cdot R_{\text{B}_1}  - \frac12\,I_{\text{B}_2}\cdot R_{\text{B}_1}\\\\
&amp;=\left(V_{\text{B}_1} - V_{\text{B}_2}\right) + \frac12\,I_{\text{B}_1}\cdot R_{\text{B}_1}- \frac12\,I_{\text{B}_1}\cdot R_{\text{B}_2} + \frac12\,I_{\text{B}_2}\cdot R_{\text{B}_1} - \frac12\,I_{\text{B}_2}\cdot R_{\text{B}_2}\\&amp;\hphantom{=\text{ } \left(V_{\text{B}_1} - V_{\text{B}_2}\right)}+ \frac12\,I_{\text{B}_1}\cdot R_{\text{B}_1} - \frac12\,I_{\text{B}_2}\cdot R_{\text{B}_1}+\frac12\,I_{\text{B}_1}\cdot R_{\text{B}_2}  - \frac12\,I_{\text{B}_2}\cdot R_{\text{B}_2}\\\\
&amp;=\left(V_{\text{B}_1} - V_{\text{B}_2}\right)+\frac12\left(I_{\text{B}_1}+I_{\text{B}_2}\right)\left(R_{\text{B}_1}-R_{\text{B}_2}\right)+\frac12\left(I_{\text{B}_1}-I_{\text{B}_2}\right)\left(R_{\text{B}_1}+R_{\text{B}_2}\right)
\end{align*}$$
Now, you just break up those three parts:
$$\begin{align*}
V_{\text{ERR}_1}&amp;=\left(R_{\text{B}_1}-R_{\text{B}_2}\right)\frac{I_{\text{B}_1}+I_{\text{B}_2}}{2}=\left(R_{\text{B}_1}-R_{\text{B}_2}\right) I_{\text{IN}_\text{BIAS}}\\\\
V_{\text{ERR}_2}&amp;=\left(R_{\text{B}_1}+R_{\text{B}_2}\right)\frac{I_{\text{B}_1}-I_{\text{B}_2}}{2}=\left(R_{\text{B}_1}-R_{\text{B}_2}\right) I_{\text{IN}_\text{OFF}}\\\\
V_{\text{ERR}_3}&amp;=V_{\text{B}_1} - V_{\text{B}_2} = V_{\text{IN}_\text{OFF}}\\\\
V_1-V_2 &amp;= V_{\text{ERR}_1} + V_{\text{ERR}_2} + V_{\text{ERR}_3}
\end{align*}$$
And, of course, \$V_1-V_2\$ has \$A_v\$ applied to it at the output differential.