1. **State the problem:** We are given three expressions involving variables $g_{moA}$, $v_x$, and $v_o$ with parameters related to resistances and voltages in an electrical circuit. 2. **Given expressions:** $$g_{moA} \left(V_{OS_{OA1}}^{-v_0}\right) + \frac{V_o^+}{r_{OA}} + \frac{V_o^+ - v_x}{R} = 0$$ $$v_x = \frac{R_G \left(g_{mBJT} R_{BJT} V_{OS_{Q1}} + V_o^+ (R_B + r_{BJT})\right)}{2 \left[(R_B + r_{BJT})(R + \frac{R_G}{2}) - R \frac{R_G}{2}\right]}$$ $$v_o = - \frac{g_{mBJT} R r_{BJT} \left(R + \frac{R_G}{2}\right) V_{OS_{Q1}} + V_o^+ R \frac{R_G}{2}}{\left[(R_B + r_{BJT})(R + \frac{R_G}{2}) - R \frac{R_G}{2}\right]}$$ 3. **Explanation:** - The first equation is a sum of currents or voltages set to zero, typical in circuit analysis. - The second and third equations define $v_x$ and $v_o$ in terms of circuit parameters. 4. **Intermediate steps:** - The denominators in $v_x$ and $v_o$ are the same, so define: $$D = (R_B + r_{BJT})(R + \frac{R_G}{2}) - R \frac{R_G}{2}$$ - Rewrite $v_x$: $$v_x = \frac{R_G \left(g_{mBJT} R_{BJT} V_{OS_{Q1}} + V_o^+ (R_B + r_{BJT})\right)}{2D}$$ - Rewrite $v_o$: $$v_o = - \frac{g_{mBJT} R r_{BJT} \left(R + \frac{R_G}{2}\right) V_{OS_{Q1}} + V_o^+ R \frac{R_G}{2}}{D}$$ 5. **Interpretation:** - These formulas relate the output voltages and intermediate voltages to the transistor parameters and resistances. - To solve for $v_o$ or $v_x$ numerically, substitute known values of parameters. 6. **Summary:** - The problem involves understanding and manipulating these expressions. - The key is recognizing the common denominator $D$ and the linear dependence on $V_o^+$ and $V_{OS_{Q1}}$. No further simplification is possible without numerical values.