Đáp án:
c. \( - 1 < x < 4\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \left\{ {0;2} \right\}\\
C = \left( {\frac{2}{{x - 2}} + \frac{{x - 1}}{{2x - {x^2}}}} \right):\left( {\frac{{x + 2}}{x} - \frac{{x - 1}}{{x - 2}}} \right)\\
= \left[ {\frac{{2x - x + 1}}{{x\left( {x - 2} \right)}}} \right]:\left[ {\frac{{{x^2} - 4 - {x^2} + x}}{{x\left( {x - 2} \right)}}} \right]\\
= \left[ {\frac{{x + 1}}{{x\left( {x - 2} \right)}}} \right].\left[ {\frac{{x\left( {x - 2} \right)}}{{x - 4}}} \right]\\
= \frac{{x + 1}}{{x - 4}}\\
b.Do:2{C^2} - 7C + 3 = 0\\
\to 2{C^2} - C - 6C + 3 = 0\\
\to C\left( {2C - 1} \right) - 3\left( {2C - 1} \right) = 0\\
\to \left[ \begin{array}{l}
C = \frac{1}{2}\\
C = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
\frac{{x + 1}}{{x - 4}} = \frac{1}{2}\\
\frac{{x + 1}}{{x - 4}} = 3
\end{array} \right.\left( {DK:x \ne 4} \right)\\
\to \left[ \begin{array}{l}
2x + 2 = x - 4\\
x + 1 = 3x - 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 6\\
2x = - 13
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 6\\
x = - \frac{{13}}{6}
\end{array} \right.\left( {TM} \right)\\
c.Để:C < 0\\
\to \frac{{x + 1}}{{x - 4}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 > 0\\
x - 4 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 < 0\\
x - 4 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 1\\
x < 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 1\\
x > 4
\end{array} \right.\left( l \right)
\end{array} \right.\\
KL: - 1 < x < 4;x \ne \left\{ {0;2} \right\}
\end{array}\)
