Tính P=3-4x-x^2 Q=2x-2-3x^2 R=2-x^2-y^2-2(x+y) S= 7-x^2-y^2-2(x+y)

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1. \(\begin{array}{l}
   P = 3 – 4x – {x^2} = 7 – \left( {{x^2} – 4x + 4} \right) = 7 – {\left( {x – 2} \right)^2}\\
   Do\,\,{\left( {x – 2} \right)^2} \ge 0 \Rightarrow 7 – {\left( {x – 2} \right)^2} \le 7\\
   \Rightarrow {P_{\max }} = 7 \Leftrightarrow x – 2 = 0 \Leftrightarrow x = 2\\
   Q = 2x – 2 – 3{x^2} = – \left( {3{x^2} – 2x} \right) – 2\\
   = – 3\left( {{x^2} – \frac{2}{3}x} \right) – 2\\
   = – 3\left( {{x^2} – 2.x.\frac{1}{3} + \frac{1}{9}} \right) – 2 + \frac{1}{3}\\
   = – \frac{5}{3} – 3{\left( {x – \frac{1}{3}} \right)^2} \le – \frac{5}{3} \Rightarrow {Q_{\max }} = – \frac{5}{3} \Leftrightarrow x = \frac{1}{3}\\
   R = 2 – {x^2} – {y^2} – 2x – 2y\\
   = – \left( {{x^2} + 2x + 1} \right) – \left( {{y^2} + 2y + 1} \right) + 4\\
   = 4 – {\left( {x + 1} \right)^2} – {\left( {y + 1} \right)^2} \le 4\\
   {R_{\max }} = 4 \Leftrightarrow x = – 1;y = – 1
   \end{array}\)

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