Đáp án:

`a)`

` P(x) = -x^2 +2x +5 = -(x^2 -2x -5) = -(x^2 -2x +1 - 6) = -(x^2 -2x+1) +6 = - (x-1)^2 +6`

Ta có ` (x-1)^2 \ge 0`

` => - (x-1)^2 \le 0 => -(x-1)^2 +6 \le 6 => P(x) \le 6`

Vậy ` GTNNNN P = 6` dấu `=` xảy ra khi ` x -1 = 0 => x = 1`

`b)`

` P = -x^2 - y^2 +4x +4y +2 = -(x^2 - 4x +4) - (y^2 -4y+4) +10 = -(x-2)^2 - (y-2)^2 +10`

Ta có ` (x-2)^2 \ge 0 ; (y-2)^2 \ge 0 `

` => - (x-2)^2 - (y-2)^2 \le 0 => -(x-2)^2 - (y-2)^2 +10 \le 10`

` => P \le 10`

Vậy ` GTNNNN P = 10` , dấu `=` xảy ra khi ` x -2 = 0 => x= 2` và ` y -2 =0 => y= 2`

$\begin{array}{l}a) \quad P = -x^2 + 2x +5\\ \to P = -x^2 + 2x - 1 + 6\\ \to P = -(x^2 - 2x + 1) + 6\\ \to P = -(x-1)^2 + 6\\ \text{Ta có:}\\ \quad (x-1)^2 \geq 0\quad \forall x\\ \to -(x-1)^2 \leq 0\quad \forall x\\ \to -(x-1)^2 + 6 \leq 6\\ Hay\quad P \leq 6\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow x - 1 = 0\Leftrightarrow x = 1\\ Vậy\,\,\min P = 6 \Leftrightarrow x = 1 b)\quad P = -x^2 -y^2 +4x + 4y +2\\ \to P = (-x^2 + 4x -4) + (-y^2 + 4y - 4) + 10\\ \to P = -(x^2 -2.2x + 2^2) - (y^2 - 2.2y + 2^2) + 10\\ \to P = -(x-2)^2 - (y-2)^2 + 10\\ \text{Ta có:}\\ \quad \begin{cases}(x-2)^2 \geq 0\quad \forall x\\(y-2)^2 \geq 0\quad \forall y\end{cases}\\ \to \begin{cases}-(x-2)^2 \leq 0\quad \forall x\\-(y-2)^2 \leq 0\quad \forall y\end{cases}\\ \to -(x-2)^2 - (y-2)^2 + 10\leq 10\\ Hay\quad P \leq 10\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \begin{cases}x - 2 = 0\\y - 2=0\end{cases}\Leftrightarrow x = y = 2\\ Vậy\,\,\min P = 10\Leftrightarrow x = y = 2 \end{array}$