Đáp án:
$\begin{array}{l}
C = \frac{{{x^2} - 3x + 3}}{{{x^2} - 2x + 1}}\left( {dk:x \ne 1} \right)\\
\Rightarrow \left( {C - 1} \right){x^2} + \left( {3 - 2C} \right)x + C - 3 = 0\\
Pt\,co\,nghiem\,x \ne 1\\
\Rightarrow \left\{ \begin{array}{l}
\left( {C - 1} \right){.1^2} + \left( {3 - 2C} \right).1 + C - 3 \ne 0\\
{\left( {3 - 2C} \right)^2} - 4\left( {C - 1} \right).\left( {C - 3} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 1 \ne 0\left( {tm} \right)\\
4{C^2} - 12C + 9 - 4{C^2} + 16C - 12 \ge 0
\end{array} \right.\\
\Rightarrow 4C \ge 3\\
\Rightarrow C \ge \frac{3}{4}\\
\Rightarrow GTNN:C = \frac{3}{4}
\end{array}$
