The output becomes 1 V, because:
When negative feedback is added to an op-amp, the input pins become identical. Meaning, whatever is the voltage present in the non-inverting input is also present in the inverting input.
The output goes as close as it can to -Vcc, because in this case it essentially becomes a comparator.
A comparator compares two input voltages and outputs a binary signal indicating which is larger. If the non-inverting (+) input is greater than the inverting (-) input, the output goes high. If the inverting input is greater than the non-inverting, the output goes low.
The output goes as close as it can to +Vcc, because it again becomes a comparator (read above).
Summarized: If plus is higher then output goes high, if minus is higer then output goes low.
I think the resistor has no real use here, it just makes it harder for the opamp to equalize the inputs? Maybe I should just disregard this circuit...
Maybe this will make it easier to understand:
The OUT 'wants' to equalize the difference between the inputs, so we can imagine INV-INPUT to be 3 V. With a bit of math we can see that OUT needs to be 3.5 V for this.
So, because there is a feedback loop the opamp will try to keep INV the same as the non-inverting input, which is at 0 V. With a bit of math we can see thast the output will be -7.5 V. $$ I_{RIN} = {V_{IN} \over R_{IN}}= {5 \over 2000}= 2.5 mA $$
$$ I_{RIN} = I_{RF} $$
$$ V_{OUT} = R_F * I_{RF} = 3000 * {5 \over 2000} = 7.5 V $$
But since it has to counteract the input voltage it has to be negative. Also, to go on: $$ gain = {V_{OUT} \over V_{IN}} = - {R_F \over R_{IN}} $$
$$ V_{INV} = 0 V $$
$$ I_{R1} = {V_1 \over R_1} = {2 \over 500} = {8 \over 2000} A $$
$$ I_{R2} = {V_2 \over R_2} = {5 \over 400} = {25 \over 2000} A $$
$$ I_{RF} = {33 \over 2000} A $$
$$ V_{RF} = I_{RF} * R_F = {33 \over 2000} * 300 = 4,95 V $$
$$ V_{OUT} = -4,95 V $$
$$ I_{R1/R2} = {{V_2 - V1} \over {R_1 + R_2}} = {1 \over 700} A \ V_{R1} = R_1 * I_{R1} = 600 * {1 \over 700} = {6 \over 7} V \ V_{IN} = V_1 + V_{R1} = 2 + {6 \over 7} = {20 \over 7} V $$
$$
$$
$$ I_{RB} = {V_{IN} \over R_B} = {{20 \over 7} \over 200} = {1 \over 70} A \ I_{RB} = I_{RA} \ V_{INV} = V_{IN} \ V_{RA} = I_{RA} * R_A = {1 \over 70} * 500 = {50 \over 70} A \ V_{OUT} = V_{INV} + V_{RA} = {20 \over 70} + {50 \over 70} = 10 V $$
When using only resistors to split the rails, the virtual ground can shift depending on the load resistors (VGND is at 2.9 V in this circuit).
This can be improved with the use of an opamp (VGND is at 4 V in this circuit).
How does it work? The opamp tries to make the inputs the same voltage, and because the positive input is steady at +4 V because of the 5k-5k voltage divider resistors V the negative will be as well.