Giải thích các bước giải:
$a)\left\{\begin{array}{l} mx-y=m+1\\ x+my=2m\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m^2x-my=m^2+m\\ x+my=2m\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m^2x-my+ x+my=m^2+m+2m\\ x+my=2m\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} (m^2+1)x+=m^2+3m\\ y=2-\dfrac{x}{m}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{m^2+3m}{m^2+1}\\ y=2-\dfrac{x}{m}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{m^2+3m}{m^2+1}\\ y=\dfrac{2m^2-m-1}{m^2+1}\end{array} \right.$
$m^2+1 >0 \ \forall \ m$ nên hệ luôn có nghiệm.
$b)A=\left(y+1\right)\left(y-2\right)+x\left(x-1\right)\\ =\left(\dfrac{2m^2-m-1}{m^2+1}+1\right)\left(\dfrac{2m^2-m-1}{m^2+1}-2\right)+\dfrac{m^2+3m}{m^2+1}\left(\dfrac{m^2+3m}{m^2+1}-1\right)\\ =\dfrac{3m^2-m}{m^2+1}.\dfrac{-m-3}{m^2+1}+\dfrac{m^2+3m}{m^2+1}.\dfrac{3m-1}{m^2+1}\\ =\dfrac{(3m^2-m)(-m-3)}{(m^2+1)^2}+\dfrac{(m^2+3m)(3m-1)}{(m^2+1)^2}\\ =\dfrac{-m(3m-1)(m+3)m+(m+3)(3m-1)}{(m^2+1)^2}\\ =0$
$\Rightarrow A$ không phụ thuộc $m.$
