Difference op amp with imperfections -- maximum dc offset
Let me see. To be able to find \$V_O\$ due to \$V_{OS}\$ we need to solve this circuit: 







simulate this circuit &ndash; Schematic created using CircuitLab

Hence, 

$$V_O = V_{OS}* \left(1+\frac{R_2}{R_1}\right)= 0.404V$$

Now the bias current: 

\$I_B = \frac{I_P+I_N}{2}\$ 

\$I_O = I_P-I_N \$ 

Where \$I_P\$ is a Non-Inverting input bias current and \$I_N\$ - Inverting Input bias current. 

From this I get: 

\$ I_P = I_B-0.5I_O = 0.45µA \$

and 

\$ I_N = I_B+0.5I_O = 0.55µA \$ 

And the circuit diagram will look like this:







simulate this circuit

And finally using superposition, we have:  

The voltage at \$V_P\$ node is 

$$ V_P = -I_P*R_3||R_4$$

and the output voltage is \$V_O' = V_p*A_v\$ and the Non-Inverting gain is \$A_v = (1+\frac{R_2}{R_1}) = \frac{R_1+R_2}{R_1}\$ therefore

$$V_O' = -I_P*R_3||R_4*\frac{R_1+R_2}{R_1}= -I_P \frac{R_3||R_4(R_1+R_2)}{R_1} = -450mV $$ 

And for the second case, we have 

$$V_O'' = I_N*R_2 = 550mV $$ 

And finally 

$$V_O = V_O'+V_O'' = -450mV + 550mV = 100mV$$ due to the input bias current. 

Therefore the maximum output DC offset voltage is: 

\$Vo = 404mV+100mV = 504mV\$ or \$304mV\$ is we assumed that the bias current is entring the op amp input.