The ideal-op-amp circuit shown has an 8 kΩ feedback resistor to the positive input, and a 1 kΩ/2 kΩ divider to its negative input. Find $V_x/I_x$ in kΩ.
The divider fixes $V_-=V_o(1/(1+2))=V_o/3$. Ideal feedback gives $V_x=V_+=V_-=V_o/3$, so $V_o=3V_x$. No current enters the positive input, hence $I_x=(V_x-V_o)/8\ \mathrm{k\Omega}=-2V_x/8\ \mathrm{k\Omega}$. Therefore $V_x/I_x=-4$ kΩ. This is a negative input resistance, with the indicated current convention.
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