Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
P = 3 - 4x - {x^2} = \left( { - 4 - 4x - {x^2}} \right) + 7 = 7 - {(x + 2)^2} \le 7 = > {P_{\max }} = 7 \Leftrightarrow x = -2\\
Q = 2x - 2 - 3{x^2} = \left( { - \frac{1}{3} + 2x - 3{x^2}} \right) - \frac{5}{3} = - \frac{5}{3} - {(\sqrt 3 x - \frac{1}{{\sqrt 3 }})^2} \le - \frac{5}{3} = > {P_{\max }} = - \frac{5}{3} \Leftrightarrow x = \frac{1}{3}
\end{array}\]
\[\begin{array}{l}
R = 2 - {x^2} - {y^2} - 2(x + y) = 4 - \left( {1 + 2x + {x^2}} \right) - (1 + 2x + {x^2}) = 4 - {(x + 1)^2} - {(y + 1)^2} \le 4 = > {R_{\max }} = 4 \Leftrightarrow \left\{ \begin{array}{l}
x = - 1\\
y = - 1
\end{array} \right.\\
S = 7 - {x^2} - {y^2} - 2(x + y) = 9 - \left( {1 + 2x + {x^2}} \right) - (1 + 2x + {x^2}) = 9 - {(x + 1)^2} - {(y + 1)^2} \le 9 = > {S_{\max }} = 9 \Leftrightarrow \left\{ \begin{array}{l}
x = - 1\\
y = - 1
\end{array} \right.
\end{array}\]
