Đáp án:
e. x=3
Giải thích các bước giải:
\(\begin{array}{l}
d.DK:x \ne 1\\
C = \dfrac{x}{{x + 3}}\\
Có:\left| C \right| > C\\
\Leftrightarrow \left[ \begin{array}{l}
C > C\left( {vô lý} \right)\\
C < - C
\end{array} \right.\\
\to 2C < 0\\
\to C < 0\\
\to \dfrac{x}{{x + 3}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x + 3 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x + 3 > 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x < - 3
\end{array} \right.\left( {vô lý} \right)\\
- 3 < x < 0
\end{array} \right.\\
\to - 3 < x < 0\\
e.{C^2} - C + 1 = {C^2} - 2C.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {C - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {C - \dfrac{1}{2}} \right)^2} \ge 0\forall C\\
\to {\left( {C - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\to Min = \dfrac{3}{4}\\
\Leftrightarrow C - \dfrac{1}{2} = 0\\
\to C = \dfrac{1}{2}\\
\to \dfrac{x}{{x + 3}} = \dfrac{1}{2}\\
\to 2x = x + 3\\
\to x = 3\left( {TM} \right)
\end{array}\)
