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