Đáp án:
$MinC = 2 \Leftrightarrow x = 2$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
C = \dfrac{{3{x^2} - 8x + 6}}{{{x^2} - 2x + 1}}\left( {DK:x \ne 1} \right)\\
\Rightarrow C - 2 = \dfrac{{3{x^2} - 8x + 6}}{{{x^2} - 2x + 1}} - 2\\
= \dfrac{{3{x^2} - 8x + 6 - 2\left( {{x^2} - 2x + 1} \right)}}{{{x^2} - 2x + 1}}\\
= \dfrac{{{x^2} - 4x + 4}}{{{x^2} - 2x + 1}}\\
= \dfrac{{{{\left( {x - 2} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}}
\end{array}$
Mà:
$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x - 2} \right)^2} \ge 0,\forall x\\
{\left( {x - 1} \right)^2} > 0,\forall x \ne 1
\end{array} \right.\\
\Rightarrow \dfrac{{{{\left( {x - 2} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}} \ge 0,\forall x \ne 1\\
\Rightarrow C - 2 \ge 0\\
\Rightarrow C \ge 2\\
\Rightarrow MinC = 2
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x - 2 = 0 \Leftrightarrow x = 2$
Vậy $MinC = 2 \Leftrightarrow x = 2$
