\[\begin{array}{l}
Bai\,\,\,3:\\
A = 5 - 8x - {x^2} = - \left( {{x^2} + 8x} \right) + 5 = - \left( {{x^2} + 8x + 16} \right) + 16 + 5\\
= - {\left( {x + 4} \right)^2} + 21 \le 21.\\
Dau\,\,\, = \,\,\,xay\,\,ra \Leftrightarrow x + 4 = 0 \Leftrightarrow x = - 4\\
\Rightarrow Max\,\,A = 21\,\,khi\,\,\,x = - 4.\\
B = - 3{x^2} - 2x + 1 = - 3\left( {{x^2} + \frac{2}{3}x} \right) + 1 = - 3\left( {{x^2} + \frac{2}{3}x + \frac{1}{9}} \right) + \frac{1}{3} + 1\\
= - 3{\left( {x + \frac{1}{3}} \right)^2} + \frac{4}{3} \le \frac{4}{3}\\
Dau\,\,\, = \,\,\,xay\,\,ra \Leftrightarrow x + \frac{1}{3} = 0 \Leftrightarrow x = - \frac{1}{3}\\
\Rightarrow Max\,\,B = \frac{4}{3}\,\,khi\,\,\,x = - \frac{1}{3}.\\
c)\,\,C = 1 + 5y - {y^2} = - \left( {{y^2} - 5y} \right) + 1 = - \left( {{y^2} - 2.\frac{5}{2}y + \frac{{25}}{4}} \right) + \frac{{25}}{4} + 1\\
= - {\left( {y - \frac{5}{2}} \right)^2} + \frac{{29}}{4} \le \frac{{29}}{4}.\\
Dau\,\,\, = \,\,\,xay\,\,ra \Leftrightarrow y - \frac{5}{2} = 0 \Leftrightarrow y = \frac{5}{2}\\
\Rightarrow Max\,\,C = \frac{{29}}{4}\,\,khi\,\,\,y = \frac{5}{2}.\\
Cau\,\,4:\\
a)\,\,\,{x^2} + x + 1 = {x^2} + 2.\frac{1}{2}x + \frac{1}{4} - \frac{1}{4} + 1 = {\left( {x + \frac{1}{2}} \right)^2} + \frac{3}{4} > \,\,\,0\,\,\forall x.\\
b)\,\,\, - 4{x^2} - 4x - 2 = - \left( {4{x^2} + 4x + 1} \right) + 1 - 2\\
= - {\left( {2x + 1} \right)^2} - 1 < \,0\,\,\,\forall x.
\end{array}\]