Another Diff. op.-Amp. circuit

$$\frac{V_X-V_R}{V_H-V_L}=\frac{R_AR_F}{R_AR_V+R_G\left({R_A+R_V}\right)}$$

$$V_D=\frac{V_CR_{11}R_3+V_XR_0R_3+V_LR_0R_{11}}{R_{11}R_3+R_0R_3+R_0R_{11}}$$
$$V_U=\frac{V_CR_{22}R_5+V_RR_4R_5+V_HR_4R_{22}}{R_{22}R_5+R_4R_5+R_4R_{22}}$$
$$FROM\ :\ V_X-V_R=\left({V_U-V_D}\right)\frac{R_F}{R_G}$$
$$IF\ :\ \cases{R_1=R_2=R_F\\ R_{10}=R_{20}=R_G\\ R_0=R_4=R_A\\ R_3=R_5=R_V}$$
$$\boxed{!\ note\ that\ the\ above\ condition\ makes\ the\ biasing\ invariant\ of\ the\ \mathbf{V_C}}$$
  $$\left({V_X-V_R}\right)\frac{R_G}{R_F}= \frac{\underline{V_C\left({R_F+R_G}\right)R_V}+\boxed{V_RR_AR_V}+V_HR_A\left({R_F+R_G}\right)}{\left({R_F+R_G}\right)R_V+R_AR_V+R_A\left({R_F+R_G}\right)}-\frac{\underline{V_C\left({R_F+R_G}\right)R_V}+\boxed{V_XR_AR_V}+V_LR_A\left({R_F+R_G}\right)}{\left({R_F+R_G}\right)R_V+R_AR_V+R_A\left({R_F+R_G}\right)}$$
$$\left({V_X-V_R}\right)\left({\frac{R_G}{R_F}+\frac{R_AR_V}{R_{\Sigma3}}}\right)=\left({V_H-V_L}\right)\frac{R_A\left({R_F+R_G}\right)}{R_{\Sigma3}}$$
$$\frac{V_X-V_R}{V_H-V_L}=\frac{R_A\left({R_F+R_G}\right)}{\cancel{R_{\Sigma3}}}·\frac{R_F·\cancel{R_{\Sigma3}}}{R_GR_{\Sigma3}+R_AR_VR_F}=$$
$$=\frac{R_A\left({R_F+R_G}\right)\frac{R_F}{R_G}}{\left({R_F+R_G}\right)R_V+R_AR_V+R_A\left({R_F+R_G}\right)+R_AR_V\frac{R_F}{R_G}}=$$
$$=\frac{R_A\cancel{\left({R_F+R_G}\right)}\frac{R_F}{R_G}}{\cancel{\left({R_F+R_G}\right)}\left({R_A+R_V}\right)+R_AR_V\frac{\cancel{\left({R_F+R_G}\right)}}{R_G}}=$$

$$=\frac{R_AR_F}{R_AR_V+R_G\left({R_A+R_V}\right)}$$

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a simulation example ::

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[Eop]