Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = - {x^2} - 2x - 2 = - \left( {{x^2} + 2x + 1} \right) - 1\\
= - {\left( {x + 1} \right)^2} - 1 \le - 1 < 0,\,\,\forall x\\
B = - {x^2} - 4x - 7 = - \left( {{x^2} + 4x + 4} \right) - 3\\
= - {\left( {x + 2} \right)^2} - 3 \le - 3 < 0,\,\,\forall x\\
C = - {x^2} - 6x - 11 = - \left( {{x^2} + 6x + 9} \right) - 2\\
= - {\left( {x + 3} \right)^2} - 2 \le - 2 < 0,\,\,\forall x\\
D = - {x^2} - x - 1 = - \left( {{x^2} + x + \frac{1}{4}} \right) - \frac{3}{4}\\
= - {\left( {x + \frac{1}{2}} \right)^2} - \frac{3}{4} \le - \frac{3}{4} < 0,\,\,\forall x\\
E = - {x^2} - 3x - 5 = - \left( {{x^2} + 3x + \frac{9}{4}} \right) - \frac{{11}}{4}\\
= - {\left( {x + \frac{3}{2}} \right)^2} - \frac{{11}}{4} \le - \frac{{11}}{4} < 0,\,\,\forall x\\
F = - 3{x^2} - 6x - 4 = - \left( {3{x^2} + 6x + 3} \right) - 1\\
= - 3.\left( {{x^2} + 2x + 1} \right) - 1 = - 3.{\left( {x + 1} \right)^2} - 1 \le - 1 < 0,\,\,\,\forall x\\
G = - 5{x^2} + 7x - 3 = - \left( {5{x^2} - 7x + \frac{{49}}{{20}}} \right) - \frac{{11}}{{20}}\\
= - 5.\left( {{x^2} - \frac{7}{5}x + \frac{{49}}{{100}}} \right) - \frac{{11}}{{20}}\\
= - 5.{\left( {x - \frac{7}{{10}}} \right)^2} - \frac{{11}}{{20}} \le - \frac{{11}}{{20}} < 0,\,\,\forall x\\
H = - 4{x^2} - 6x - 4 = - \left( {4{x^2} + 6x + \frac{9}{4}} \right) - \frac{7}{4}\\
= - {\left( {2x + \frac{3}{2}} \right)^2} - \frac{7}{4} \le - \frac{7}{4} < 0,\,\,\forall x\\
K = - \frac{1}{2}{x^2} - x - 1 = - \left( {\frac{1}{2}{x^2} + x + \frac{1}{2}} \right) - \frac{1}{2}\\
= - \frac{1}{2}\left( {{x^2} + 2x + 1} \right) - \frac{1}{2} = - \frac{1}{2}{\left( {x + 1} \right)^2} - \frac{1}{2} \le - \frac{1}{2} < 0,\,\forall x
\end{array}\)
