Giải thích các bước giải:
\(\begin{array}{l}
*)\\
B = 3 - \sqrt {{x^2} - 8x + 16} = 3 - \sqrt {{{\left( {x - 4} \right)}^2}} = 3 - \left| {x - 4} \right|\\
\left| {x - 4} \right| \ge 0,\,\,\forall x \Rightarrow 3 - \left| {x - 4} \right| \le 3,\,\,\,\forall x\\
\Rightarrow {B_{\max }} = 3 \Leftrightarrow \left| {x - 4} \right| = 0 \Leftrightarrow x = 4\\
*)\\
\left| a \right| + \left| b \right| \ge \left| {a + b} \right|\\
P = \sqrt {{x^2} + 4x + 4} + \sqrt {{x^2} - 6x + 9} \\
= \sqrt {{{\left( {x + 2} \right)}^2}} + \sqrt {{{\left( {x - 3} \right)}^2}} \\
= \left| {x + 2} \right| + \left| {x - 3} \right|\\
= \left| {x + 2} \right| + \left| {3 - x} \right| \ge \left| {\left( {x + 2} \right) + \left( {3 - x} \right)} \right| = 5\\
\Rightarrow {P_{\min }} = 5 \Leftrightarrow \left( {x + 2} \right)\left( {3 - x} \right) \ge 0 \Leftrightarrow \left( {x + 2} \right)\left( {x - 3} \right) \le 0 \Leftrightarrow - 2 \le x \le 3\\
*)\\
A = 6 - 2x - \sqrt {9 - 6x + {x^2}} \\
= 6 - 2x - \sqrt {{{\left( {3 - x} \right)}^2}} \\
= 6 - 2x - \left| {3 - x} \right|\\
TH1:\,\,\,\,\,x \le 3 \Rightarrow 3 - x \ge 0\\
\Rightarrow A = 6 - 2x - \left( {3 - x} \right) = 3 - x\\
TH2:\,\,\,\,\,\,x > 3 \Rightarrow 3 - x < 0\\
\Rightarrow A = 6 - 2x + \left( {3 - x} \right) = 9 - 3x\\
b,\\
x = 5 > 3 \Rightarrow A = 9 - 3x = 9 - 3.5 = - 6\\
c,\\
A = 0 \Leftrightarrow \left[ \begin{array}{l}
3 - x = 0\\
9 - 3x = 0
\end{array} \right. \Leftrightarrow x = 3
\end{array}\)
