Đáp án:
a) MinA=2020
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\dfrac{3}{2} \ge x\\
A = 2020 + \sqrt {3 - 2x} \\
Do:\sqrt {3 - 2x} \ge 0\\
\to 2020 + \sqrt {3 - 2x} \ge 2020\\
\to MinA = 2020\\
\Leftrightarrow 3 - 2x = 0\\
\to x = \dfrac{3}{2}\\
b)C = 2021 - \sqrt {{x^2} - 1} \\
DK:\left[ \begin{array}{l}
x \ge 1\\
x \le - 1
\end{array} \right.\\
Do:\sqrt {{x^2} - 1} \ge 0\\
\to - \sqrt {{x^2} - 1} \le 0\\
\to 2021 - \sqrt {{x^2} - 1} \le 2021\\
\to Max = 2021\\
\Leftrightarrow {x^2} - 1 = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\\
D = 6 + \sqrt {4x - {x^2} - 2} \\
= 6 + \sqrt { - \left( {{x^2} - 4x + 2} \right)} \\
= 6 + \sqrt { - \left( {{x^2} - 4x + 4 - 2} \right)} \\
= 6 + \sqrt { - {{\left( {x - 2} \right)}^2} + 2} \\
Do: - {\left( {x - 2} \right)^2} \le 0\forall x\\
\to - {\left( {x - 2} \right)^2} + 2 \le 2\\
\to \sqrt { - {{\left( {x - 2} \right)}^2} + 2} \le \sqrt 2 \\
\to 6 + \sqrt { - {{\left( {x - 2} \right)}^2} + 2} \le 6 + \sqrt 2 \\
\to Max = 6 + \sqrt 2 \\
\Leftrightarrow x = 2
\end{array}\)
