Đáp án:
b) \(Max = \dfrac{{11}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = 4x - {x^2} + 3\\
= - \left( {{x^2} - 4x - 3} \right)\\
= - \left( {{x^2} - 4x + 4 - 7} \right)\\
= - {\left( {x - 2} \right)^2} + 7\\
Do:{\left( {x - 2} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 2} \right)^2} \le 0\\
\to - {\left( {x - 2} \right)^2} + 7 \le 7\\
\to Max = 7\\
\Leftrightarrow x - 2 = 0\\
\Leftrightarrow x = 2\\
b)B = 2x - 2{x^2} + 5\\
= - \left( {2{x^2} - 2x - 5} \right)\\
= - \left( {2{x^2} - 2.x\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} - \dfrac{{11}}{2}} \right)\\
= - {\left( {x\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{{11}}{2}\\
Do:{\left( {x\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to - {\left( {x\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} \le 0\\
\to - {\left( {x\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{{11}}{2} \le \dfrac{{11}}{2}\\
\to Max = \dfrac{{11}}{2}\\
\Leftrightarrow x\sqrt 2 - \dfrac{1}{{\sqrt 2 }} = 0\\
\Leftrightarrow x = \dfrac{1}{2}
\end{array}\)
