Đáp án:
a) $MinA = - 2 \Leftrightarrow x = 4$
b) $MinB = 6 \Leftrightarrow x = 4$
c) $MinC = \dfrac{{ - 1}}{4} \Leftrightarrow x = \dfrac{1}{4}$
d) $MinD = 1 + \sqrt 3 \Leftrightarrow x = 1$
Giải thích các bước giải:
$a)A = \sqrt {x - 4} - 2\left( {DK:x \ge 4} \right)$
Ta có:
$\begin{array}{l}
\sqrt {x - 4} \ge 0,\forall x \ge 4\\
\Rightarrow A = \sqrt {x - 4} - 2 \ge - 2,\forall x \ge 4
\end{array}$
Dấu bằng xảy ra$ \Leftrightarrow \sqrt {x - 4} = 0 \Leftrightarrow x = 4$
Vậy $MinA = - 2 \Leftrightarrow x = 4$
$\begin{array}{l}
b)B = x - 4\sqrt x + 10\left( {DK:x \ge 0} \right)\\
= \left( {x - 4\sqrt x + 4} \right) + 6\\
= {\left( {\sqrt x - 2} \right)^2} + 6
\end{array}$
Mà:
$\begin{array}{l}
{\left( {\sqrt x - 2} \right)^2} \ge 0,\forall x \ge 0\\
\Rightarrow {\left( {\sqrt x - 2} \right)^2} + 6 \ge 6,\forall x \ge 0\\
\Rightarrow B \ge 6,\forall x \ge 0
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow {\left( {\sqrt x - 2} \right)^2} = 0 \Leftrightarrow \sqrt x - 2 = 0 \Leftrightarrow x = 4$
Vậy $MinB = 6 \Leftrightarrow x = 4$
$\begin{array}{l}
c)C = x - \sqrt x \left( {DK:x \ge 0} \right)\\
= x - 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4}\\
= {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4}
\end{array}$
Mà:
$\begin{array}{l}
{\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \ge 0,\forall x \ge 0\\
\Rightarrow {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge \dfrac{{ - 1}}{4},\forall x \ge 0\\
\Rightarrow C \ge \dfrac{{ - 1}}{4},\forall x \ge 0
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow \sqrt x - \dfrac{1}{2} = 0 \Leftrightarrow x = \dfrac{1}{4}$
Vậy $MinC = \dfrac{{ - 1}}{4} \Leftrightarrow x = \dfrac{1}{4}$
$\begin{array}{l}
d)D = \sqrt {{x^2} - 2x + 4} + 1\\
= \sqrt {{{\left( {x - 1} \right)}^2} + 3} + 1
\end{array}$
Mà:
$\begin{array}{l}
{\left( {x - 1} \right)^2} \ge 0,\forall x\\
\Rightarrow {\left( {x - 1} \right)^2} + 3 \ge 3,\forall x\\
\Rightarrow \sqrt {{{\left( {x - 1} \right)}^2} + 3} + 1 \ge 1 + \sqrt 3 ,\forall x\\
\Rightarrow D \ge 1 + \sqrt 3
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow {\left( {x - 1} \right)^2} = 0 \Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1$
Vậy $MinD = 1 + \sqrt 3 \Leftrightarrow x = 1$
