A small hint.
Practical Op-Amps have a parameter called bias current. This is the small current that the Op Amp draws from it's input pins.
There are opamps that actually GIVE current out of their input pins. Jfet opamps like the tl082 are notorious for that.
this current can , if you use large resistance values, create enough offset to mess things up.
In precision circuitry you will often see an extra resistor between the feedback node and the actual input.
------r2----
| |
--r1--+--r3--|-\ |
| >--+---
-r4--|+/
|
GND
R3 and R4 serve no purpose in an 'ideal opamp'. Input impedance is infinite...
In a real opamp there Is an input current. By making R3 = R4 the current drawn , or delivered, by the inputs creates a voltage drop across R3 and R4. Since both are the same value this drop is equal and thus becomes a common-mode signal, which is rejected by the opamp.
An opamp is only interested in differentials , anything common is rejected. So that's how you solve the problem.
Now, there is more. This assumption is only true IF the node formed by R1 and R2 is low impedant ! otherwise it is possible to still pull that one out of whack since that current going into or out of the inut may still influence that node.
If r1 and r2 are in the megaohm range you don't stand a chance of getting the bias cancelled, unless you use active cancelling ( by using an additional opamp to sense the drop across R4 and feed it back into the node. Or you add on a matched pair of fets or mosfets at the input to drop this bias current. That is for example youw you build an electrometer that can measure femtoamperes. you can't do that with a normal opamp as it may be faulting the measurement by injecting or drawing simply too much current itself...
Most opamps have bias current in the microamp range , some go all the way down to femto ampere.
Where is this important ? well, let's say you have a sensor with an output impedance of 10 megaohm. The sensor delivers a voltage between 1 and 100 millivolts. The opamp is supplied with a 5 volts power supply.
If your opamp has a input bias current of 1 uampere ... 1 microampere through 10 megaohm is ... 10 volt ! You can kiss your signal goodbye... the opamp is already well outside its operating range ( it only has 5 volts to work with ) so the input is clipping at 5 volts ( assuming it's a rail to rail input opamp and then still it would be like 4.7 volt . if it's a classic, non rail to rail input, opamp and the pump happens to be topside it may clip at 3 volts ... ).
So that circuit will never work. moreover , you are overloading the output of the sensor with a high voltage and you may destroy it !
Think this is exagerated example ? the carbon monoxide sensor in your house's fire alarms are exactly that. They are an electrochemical cell that delivers a few millivolts across a 1 megaohm resistor. Hook up the wrong kind of opamp and your system doesn't stand a snowball's chance in hell to work. As an extra: apply more than 100 millivolts externally and you kill the sensor electrode.
So yes, input bias current is an extremely important factor in certain designs. in others ? meh , who cares.
And there is no amount of maths that is going to teach you that. You need to know what to look for and what you are working with. Sure, you will use math to see if you fall within range. But it's easier if you know what to look for without having to calculate everything everytime.