Đáp án:
$\begin{array}{l}
a)A = {x^2} - 6x + \frac{9}{4}\\
= {x^2} - 2.3x + 9 - \frac{{27}}{4}\\
= {\left( {x - 3} \right)^2} - \frac{{27}}{4} \ge - \frac{{27}}{4}\\
GTNN: - \frac{{27}}{4},khi:x = 3\\
b)B = - {x^2} - {y^2} + 4x - 2y + 7\\
= - \left( {{x^2} - 4x + 4 + {y^2} + 2y + 1} \right) + 12\\
= - {\left( {x - 2} \right)^2} - {\left( {y + 1} \right)^2} + 12 \le 12\\
GTLN:12,khi;x = 2:y = - 1\\
c)Do:a + b + c = 0\\
\Leftrightarrow a = - \left( {b + c} \right)\\
\Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc\\
= {\left( { - b - c} \right)^3} + {b^3} + {c^3} + 3\left( {b + c} \right).bc\\
= - {b^3} - 3{b^2}c - 3b{c^2} - {c^3} + {b^3} + {c^3} + 3{b^3}c + 3b{c^2}\\
= 0\\
Vay\,{a^3} + {b^3} + {c^3} = 3abc
\end{array}$