Đáp án:
b. \(Max = 6\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Do:{x^2} \ge 0\forall x \in R\\
\to - 2{x^2} \le 0\\
\to 3 - 2{x^2} \le 3\\
\to \sqrt {3 - 2{x^2}} \le \sqrt 3 \\
\to Max = \sqrt 3 \\
\Leftrightarrow x = 0\\
b.B = 5 + \sqrt { - \left( {4{x^2} + 4x} \right)} \\
= 5 + \sqrt { - \left( {4{x^2} + 4x + 1 - 1} \right)} \\
= 5 + \sqrt {1 - {{\left( {2x + 1} \right)}^2}} \\
Do:{\left( {2x + 1} \right)^2} \ge 0\forall x \in R\\
\to - {\left( {2x + 1} \right)^2} \le 0\\
\to 1 - {\left( {2x + 1} \right)^2} \le 1\\
\to \sqrt {1 - {{\left( {2x + 1} \right)}^2}} \le \sqrt 1 \\
\to 5 + \sqrt {1 - {{\left( {2x + 1} \right)}^2}} \le 6\\
\to Max = 6\\
\Leftrightarrow 2x + 1 = 0\\
\Leftrightarrow x = - \dfrac{1}{2}
\end{array}\)
