Đáp án:
\(\begin{array}{l}
a)Max = 20\\
b)Max = \dfrac{1}{4}\\
c)Max = 7\\
d)Max = 21\\
e)Max = 5
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a) - {x^2} + 6x - 11\\
= - \left( {{x^2} - 6x + 11} \right)\\
= - \left( {{x^2} - 6x + 9 - 20} \right)\\
= - {\left( {x - 3} \right)^2} + 20\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 3} \right)^2} \le 0\\
\to - {\left( {x - 3} \right)^2} + 20 \le 20\\
\to Max = 20\\
\Leftrightarrow x = 3\\
b)x - {x^2} = - \left( {{x^2} - x} \right)\\
= - \left( {{x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4}} \right)\\
= - {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4}\\
Do:{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0\forall x\\
\to - {\left( {x - \dfrac{1}{2}} \right)^2} \le 0\\
\to - {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
\to Max = \dfrac{1}{4}\\
\Leftrightarrow x = \dfrac{1}{2}\\
c)4x - {x^2} + 3 = - \left( {{x^2} - 4x - 3} \right)\\
= - \left( {{x^2} - 4x + 4 - 7} \right)\\
= - {\left( {x - 2} \right)^2} + 7\\
Do:{\left( {x - 2} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 2} \right)^2} \le 0\\
\to - {\left( {x - 2} \right)^2} + 7 \le 7\\
\to Max = 7\\
\Leftrightarrow x = 2\\
d)5 - 8x - {x^2} = - \left( {{x^2} + 8x - 5} \right)\\
= - \left( {{x^2} + 8x + 16 - 21} \right)\\
= - {\left( {x + 4} \right)^2} + 21\\
Do:{\left( {x + 4} \right)^2} \ge 0\forall x\\
\to - {\left( {x + 4} \right)^2} \le 0\\
\to - {\left( {x + 4} \right)^2} + 21 \le 21\\
\to Max = 21\\
\Leftrightarrow x = - 4\\
e)4x - {x^2} + 1 = - \left( {{x^2} - 4x - 1} \right)\\
= - \left( {{x^2} - 4x + 4 - 5} \right)\\
= - {\left( {x - 2} \right)^2} + 5\\
Do:{\left( {x - 2} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 2} \right)^2} \le 0\\
\to - {\left( {x - 2} \right)^2} + 5 \le 5\\
\to Max = 5\\
\Leftrightarrow x = 2
\end{array}\)