$A, a^2x + x = 2a^4 – 2 ⇔ x(a^2 + 1) = 2(a^4 – 1)$
$X=\frac{2.(a^4-1)}{a^2+1}=$ $\frac{2.(a^2-1).(a^2+1)}{a^2+1}=2.(a^2-1)$
$B, a^2x + 3ax + 9 = a^2 ⇔ ax(a + 3) = a^2 – 9$
$⇔x=\frac{a^2-9}{a.(a+3)}$ $=\frac{(a+3).(a-3)}{a.(a+3)}=$ $\frac{3-a}{a}$$(với$.$a ≠0;a ≠-3)$