Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
A = 1 + \sqrt {4x - {x^2} - 2} \\
= 1 + \sqrt { - \left( {{x^2} - 4x + 4} \right) + 2} \\
= 1 + \sqrt {2 - {{\left( {x - 2} \right)}^2}} \\
{\left( {x - 2} \right)^2} \ge 0,\,\,\forall x \Rightarrow 2 - {\left( {x - 2} \right)^2} \le 2,\,\,\forall x\\
\Rightarrow A = 1 + \sqrt {2 - {{\left( {x - 2} \right)}^2}} \le 1 + \sqrt 2 ,\,\,\,\forall x\\
\Rightarrow {A_{\max }} = 1 + \sqrt 2 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2\\
d,\\
B = 1 - \sqrt {{x^2} - 2x + 2} \\
= 1 - \sqrt {\left( {{x^2} - 2x + 1} \right) + 1} \\
= 1 - \sqrt {{{\left( {x - 1} \right)}^2} + 1} \\
{\left( {x - 1} \right)^2} \ge 0,\,\,\forall x \Rightarrow {\left( {x - 1} \right)^2} + 1 \ge 1,\,\,\,\forall x\\
\Rightarrow B = 1 - \sqrt {{{\left( {x - 1} \right)}^2} + 1} \le 1 - 1 = 0,\,\,\,\forall x\\
\Rightarrow {B_{\max }} = 0 \Leftrightarrow {\left( {x - 1} \right)^2} = 0 \Leftrightarrow x = 1
\end{array}\)
