$C=9x^2+12x+6$
$C=(9x^2+12x+4)+2$
$C=(3x+2)^2+2$
Vì $(3x+2)^2\ge 0 \; \forall x\in \mathbb{R}$
$⇒ (3x+2)^2+2\ge 2 \; \forall x\in \mathbb{R}$
Vậy $\min C=2$ khi $3x+2=0 ⇔ x=-\dfrac23$
$D=x^2+x+12$
$D=\left(x^2+x+\dfrac14\right) + \dfrac{47}{4}$
$D=\left(x+\dfrac12\right)^2+ \dfrac{47}{4}$
Vì $\left(x+\dfrac12\right)^2\ge 0 \; \forall x\in \mathbb{R}$
$⇒ \left(x+\dfrac12\right)^2+ \dfrac{47}{4}\ge \dfrac{47}{4} \; \forall x\in \mathbb{R}$
Vậy $\min D=\dfrac{47}{4}$ khi $x+\dfrac12=0 ⇔ x=-\dfrac12$
$E=y^4+y^2-49$
Vì $\begin{cases}y^4\ge 0\\y^2\ge 0\end{cases}$
$⇒ y^4+y^2-49 \ge -49$
Vậy $\min E=-49$ khi $y=0$
