Đáp án:
\(MaxA = 5\)
Giải thích các bước giải:
\(\begin{array}{l}
A = 5 - \sqrt {3x + 2} \\
Do:\sqrt {3x + 2} \ge 0\forall x \ge - \dfrac{2}{3}\\
\to - \sqrt {3x + 2} \le 0\\
\to 5 - \sqrt {3x + 2} \le 5\\
\to MaxA = 5\\
\Leftrightarrow 3x + 2 = 0\\
\Leftrightarrow x = - \dfrac{2}{3}\\
B = \sqrt {{x^2} - 2x + 1 + 8} \\
= \sqrt {{{\left( {x - 1} \right)}^2} + 8} \\
Do:{\left( {x - 1} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x - 1} \right)^2} + 8 \ge 8\\
\to \sqrt {{{\left( {x - 1} \right)}^2} + 8} \ge 2\sqrt 2 \\
\to MinB = 2\sqrt 2 \\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1\\
C = 6 + \sqrt { - \left( {{x^2} - 4x + 2} \right)} \\
= 6 + \sqrt { - {{\left( {x - 2} \right)}^2} + 2} \\
Do:{\left( {x - 2} \right)^2} \ge 0\forall x\\
\to - {\left( {x - 2} \right)^2} \le 0\\
\to - {\left( {x - 2} \right)^2} + 2 \le 2\\
\to \sqrt { - {{\left( {x - 2} \right)}^2} + 2} \le \sqrt 2 \\
\to 6 + \sqrt { - {{\left( {x - 2} \right)}^2} + 2} \le 6 + \sqrt 2 \\
\to MaxC = 6 + \sqrt 2 \\
\Leftrightarrow x - 2 = 0\\
\Leftrightarrow x = 2\\
D = \sqrt {x - 1 + 2\sqrt {x - 1} .1 + 1} + \sqrt {x - 1 - 2\sqrt {x - 1} .1 + 1} + 9\\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} + 9\\
= \left| {\sqrt {x - 1} + 1} \right| + \left| {\sqrt {x - 1} - 1} \right| + 9\\
= \sqrt {x - 1} + 1 + \sqrt {x - 1} - 1 + 9\left( {DK:x \ge 1} \right)\\
= 2\sqrt {x - 1} + 9\\
Do:2\sqrt {x - 1} \ge 0\forall x \ge 1\\
\to 2\sqrt {x - 1} + 9 \ge 9\\
\to MinD = 9\\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1
\end{array}\)
