Giải thích các bước giải:
$y=\dfrac{x+a}{x^2+1+2a}$
$\rightarrow yx^2+y+2ay=x+a$
$\rightarrow yx^2-x+(y+2ay-a)=0$
$\rightarrow\Delta = 1-4y(y+2ay-a)\ge 0$
$\rightarrow -4y^2(2a+1)+4ay+1=0$ có 2 nghiệm $M, m$ thỏa mãn $3M+7m=0$
$\rightarrow M+m=\dfrac{a}{2a+1}, M.m=\dfrac{-1}{4(2a+1)}$
$\rightarrow m=\dfrac{-3}{4}(M+m)=\dfrac{-3a}{4(2a+1)}$
$\rightarrow M=-\dfrac{7}{3}m=\dfrac{7a}{4(2a+1)}$
$\rightarrow \dfrac{-3a}{4(2a+1)}.\dfrac{7a}{4(2a+1)}=\dfrac{-1}{4(2a+1)}$
$\rightarrow a=\dfrac{2}{3}$
$\rightarrow A$