I want to find the transfer function of this circuit, and I start by finding the voltage at the positive input of op-amp 1. If my understanding is correct, voltage division gives If the op-amp is assumed to be ideal, this will also be the voltage at the negative input, and since there is no voltage drop on the line between the negative input and the uotput, the output of this op-amp has no gain. Is this correct so far?
The first opamps have a very high input impedance so they don't load the filters at their inputs, a very low output impedance so the mixing resistors do not load them down and have a gain of exactly 1.
Yes you are right. Find the input of op amp 2 too. Op amps 1 and 2 are voltage followers, so their outputs are the same as their inputs (almost for real ones). Op amp 3 is a summing amplifier thus add the output voltages of 1 and 2 and multiply them by the appropriate gain.