Buffered voltage divider vs inverting amplifier
Besides the valuable practical advice above, it is interesting to reveal the "philosophy" behind these dual circuit implementations of a voltage attenuator. I will try to show that there is a close connection between them and the inverting version is derivative of the non-inverting one.

In the conceptual picture below, I have visualized the voltages with vertical bars in red, which height is proportional to the voltage value (in association with a water column). Positive voltages are above and negative voltages - below the zero voltage line (the ground). The currents are represented by closed paths in green. 



1. Non-inverting voltage divider. The idea of the classic voltage divider is simple - to pass the current through a network of resistors in series and take some of the voltage drops as an output (Fig. 1 on the left). As we prefer a grounded output, we have chosen \$V_{R2}\$ to be output voltage. The output resistance \$R_2\$ is a part of the total resistance \$R_1 + R_2\$; the output voltage \$V_{R2}\$ is the same part of the input voltage \$V_{IN}\$.

To make the transfer ratio simply \$K = R_2/R_1\$, we would desire to set the current by applying \$V_{IN}\$ directly to \$V_{R}\$ and getting \$V_{R2}\$ as output voltage... but it is impossible in this simple configuration. We can apply the input voltage to \$R_1\$ only through \$R_2\$; hence the more complex transfer ratio \$K = R_2/(R_1 + R_2\$). The current will be also less than \$V_{R1}/R_1\$.

2. Inverting voltage divider. But we can do this "magic" by means of electronics. In addition to the classic view of introducing the inverting configuration, I will use a few more unconventional perspectives to show the clever idea behind it.

- Reverse voltage divider. If we examine carefully the second circuit above, we will see that the op-amp OA2 adds its output voltage \$V_{OUT}\$ to the input voltage. To show this, I have drawn the conceptual Fig. 2 on the right where the op-amp is represented by a variable voltage source \$V_{OUT}\$ driven by the output voltage of the voltage divider  - the common point at the op-amp inverting input (aka virtual ground). It is seen that both voltage sources \$V_{IN}\$ and \$V_{OUT}\$ are connected (through the common ground) in series and unidirectionally so their voltages are added. The total voltage \$V_{IN} + V_{OUT}\$ now serves as a new input voltage for the same voltage divider.

In this negative feedback circuit, the op-amp changes its own output voltage and accordingly, the total input voltage of the voltage divider, to make the voltage drop across \$R_1\$ equal to the input voltage \$V_{IN}\$. To do so, it "goes down below zero," ie, its output voltage is negative. But where is the output voltage here?

It seems paradoxical but why not use the total input voltage as output voltage? Then it will be a "reverse voltage divider" which output is input and its input - output. In contrast to the humble passive voltage divider, its gain will be more than 1... and it will be another ubiquitous op-amp circuit - non-inverting amplifier. So we can think of the non-inverting amplifier as of a "reverse voltage divider". This is a unique property of the negative feedback - to reverse devices (swap their inputs and outputs).

But we are talking here about the inverting version... So, where can we take another output voltage from? We can use the voltage drop across \$R_2\$ as an output so the gain will be the tidy \$K = R_2/R_1\$ ratio. But there are two drawbacks here: first, \$V_{R2}\$ is floating; second, the load in parallel to \$R_2\$ will divert some current thus introducing some error. What is left for us to do then? Eureka! Instead the original voltage \$V_{R2}\$, we can use its negative "copy" \$V_{OUT}\$ as a grounded output. The gain will be tidy but negative \$K = -R_2/R_1\$.

- Voltage compensation. We can look at another perspective at the humble voltage divider (Fig. 1 in the left) if we realize the fact the output voltage drop \$V_{R2}\$ is responsible for the reduction of input current; so, we conclude, it is undesired. Then let's destroy (compensate, neutralize) it by equivalent voltage \$V_{OUT}\$ (Fig. 2) subtracting it in a series manner (the two voltages are connected contrary in series). To do it, the op-amp continuously "observes" (by its inverting input) the voltage \$V_{(-)}\$ at the upper end of \$V_{R2}\$ and changes its output voltage \$V_{OUT}\$ until zeroes it (the so-called virtual ground appears there).

We are proud that we have successfully eliminated the undesired voltage... But now we remember this was our output voltage.. we needed it... but we destroyed it... Then what do we do? Where to take an output from? Fortunately, there is a "copy" of this voltage... inverted but buffered and grounded... and we take it as output voltage.

- Negative resistance. We have solved the contradiction above (\$V_{R2}\$ was both useful and undesired) in terms of voltages. We can do it in terms of resistances if we realize the fact the op-amp output behaves as an equivalent negative "resistor" with resistance \$-R_2\$. Why? In contrast to the ordinary "positive" resistance \$R_2\$ that subtracts its voltage drop \$V_{R2}\$ from the input voltage, the op-amp adds the same voltage to the input voltage.

So, in this configuration, two equal but opposite resistors are connected in series... and the negative resistance neutralizes the positive resistance. The result is zero resistance; the network of both elements in series behaves as a piece of wire. Only the other two elements - \$V_{IN}\$ and \$R_1\$ remain in the circuit... and only they determine the current \$I = V_{IN}/R_1\$...

- Resistor summer. Also, we can think of this arrangement as of a "shifted voltage divider" since another voltage source is inserted between the lower end of \$R_2\$ and ground. And what is better, we can recognize another ubiquitous passive circuit - the resistor voltage summer. It adds two grounded (single-ended) voltages (here subtracts as they are opposite) and the result is also single-ended voltage referenced to ground. We can find it by applying the superposition principle thinking of the network once as a voltage divider driven from the left and then from the right.

The op-amp does the same as above - it keeps the summer output equal to zero. So the result is the same - \$K = -R_2/R_1\$.

- Tug of war. And finally, if you like analogies, here is a mechanical one. You can think of this arrangement as of "tug of war" where the input voltage source "pulls up" the common point between resistors while the op-amp output "pulls" it down. They do it through the resistors... and none of them succeeds... so the point does not move.

(Indeed, this story has become quite lengthy... but that is the price of understanding...)