Đáp án:
$\begin{array}{l}
a)MinA = 5 \Leftrightarrow x = 1\\
b)MinB = 3 \Leftrightarrow x = - 1\\
c)MinC = 5 \Leftrightarrow - 2 \le x \le 3\\
d)MinD = 2 \Leftrightarrow - 1 \le x \le 1
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
a)A = \sqrt {2{x^2} - 4x + 27} \\
= \sqrt {2\left( {{x^2} - 2x + 1} \right) + 25} \\
= \sqrt {2{{\left( {x - 1} \right)}^2} + 25} \\
\ge \sqrt {25} \left( {do:{{\left( {x - 1} \right)}^2} \ge 0,\forall x} \right)\\
= 5\\
\text{Dấu bằng xảy ra} \Leftrightarrow {\left( {x - 1} \right)^2} = 0 \Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1\\
\text{Vậy} MinA = 5 \Leftrightarrow x = 1\\
b)B = \sqrt {3{x^2} + 6x + 12} \\
= \sqrt {3\left( {{x^2} + 2x + 1} \right) + 9} \\
= \sqrt {3{{\left( {x + 1} \right)}^2} + 9} \\
\ge \sqrt 9 \left( {do:{{\left( {x + 1} \right)}^2} \ge 0,\forall x} \right)\\
= 3\\
\text{Dấu bằng xảy ra} \Leftrightarrow {\left( {x + 1} \right)^2} = 0 \Leftrightarrow x + 1 = 0 \Leftrightarrow x = - 1\\
\text{Vậy} MinB = 3 \Leftrightarrow x = - 1\\
c)C = \sqrt {{x^2} - 6x + 9} + \sqrt {{x^2} + 4x + 4} \\
= \sqrt {{{\left( {x - 3} \right)}^2}} + \sqrt {{{\left( {x + 2} \right)}^2}} \\
= \left| {x - 3} \right| + \left| {x + 2} \right|\\
= \left| {3 - x} \right| + \left| {x + 2} \right|\\
\ge \left| {3 - x + x + 2} \right|\\
= 5\\
\text{Dấu bằng xảy ra} \Leftrightarrow \left( {3 - x} \right)\left( {x + 2} \right) \ge 0\\
\Leftrightarrow - 2 \le x \le 3\\
\text{Vậy} MinC = 5 \Leftrightarrow - 2 \le x \le 3\\
d)D = \sqrt {{x^2} - 2x + 1} + \sqrt {{x^2} + 2x + 1} \\
= \sqrt {{{\left( {x - 1} \right)}^2}} + \sqrt {{{\left( {x + 1} \right)}^2}} \\
= \left| {x - 1} \right| + \left| {x + 1} \right|\\
= \left| {1 - x} \right| + \left| {x + 1} \right|\\
\ge \left| {1 - x + x + 1} \right|\\
= 2\\
\text{Dấu bằng xảy ra} \Leftrightarrow \left( {1 - x} \right)\left( {x + 1} \right) \ge 0\\
\Leftrightarrow - 1 \le x \le 1\\
\text{Vậy} MinD = 2 \Leftrightarrow - 1 \le x \le 1
\end{array}$
