Cho hệ phương trình mx-y=m^2 và 2x+my=-m^3+2m+2. Trong mọi trường hợp

* Hướng dẫn giải

$$\begin{gathered} \left\{\begin{array}{l}mx-y=m^2\\ 2x+my=-m^3+2m+2\end{array}\right. \\ \iff \left\{\begin{array}{l}y=mx-m^2\\ 2x+m\left(mx-m^2\right)=-m^3+2m+2\end{array}\right. \\ \iff \left\{\begin{array}{l}y=mx-m^2\\ x\left(m^2+2\right)=2m+2\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{{\displaystyle 2m+2}}{{\displaystyle m^2+2}}\\ y=m.\frac{{\displaystyle 2m+2}}{{\displaystyle m^2+2}}-m^2\end{array}\right. \\ \iff \left\{\begin{array}{l}x=\frac{{\displaystyle 2m+2}}{{\displaystyle m^2+2}}\\ y=\frac{{\displaystyle -m^4+2m}}{{\displaystyle m^2+2}}\end{array}\right. \end{gathered}$$

(vì $m^2+2>0;\forall m$ )

Suy ra $x–y=\frac{m^4+2}{m^2+2}$  

Đáp án: C