Đáp án:
$\begin{array}{l}
a)A = {x^2} + 12x + 39\\
= {x^2} + 2.6.x + 36 + 3\\
= {\left( {x + 6} \right)^2} + 3 > 0\\
b)B = 9{x^2} - 12x\\
= 9{x^2} - 2.3.2.x + 4 - 4\\
= {\left( {3x - 2} \right)^2} - 4 \ge - 4
\end{array}$
Nên B ko thể luôn dương
$\begin{array}{l}
c)C = 2{x^2} - 6x + 8\\
= 2\left( {{x^2} - 3x} \right) + 8\\
= 2.\left( {{x^2} - 2.x.\dfrac{3}{2} + \dfrac{9}{4}} \right) - 2.\dfrac{9}{4} + 8\\
= 2.{\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{7}{2} \ge \dfrac{7}{2} > 0\\
Vậy\,C > 0\\
d)D = 9{x^2} - 6xy + 2{y^2} + 1\\
= 9{x^2} - 6xy + {y^2} + {y^2} + 1\\
= {\left( {3x - y} \right)^2} + {y^2} + 1 \ge 1 > 0\\
Vậy\,D > 0\\
e)C = {x^2} - 6x + {y^2} + 10y + 35\\
= {x^2} - 6x + 9 + {y^2} + 10y + 25 + 1\\
= {\left( {x - 3} \right)^2} + {\left( {y + 5} \right)^2} + 1 \ge 1 > 0\\
Vậy\,C > 0
\end{array}$
