Tìm \(x, \) biết :
a) \(4{x^2} - 1 - \left( {1 - 2x} \right) \left( {x + 2} \right) = 0; \) b) \( \frac{{3x - {x^2}}}{{{x^2} - 9}} = 0. \)
A.\(\begin{array}{l}{\rm{a)}}\,\,S = \left\{ {\frac{1}{2}; - 1} \right\}\\{\rm{b)}}\,\,x = 0\end{array}\)
B.\(\begin{array}{l}{\rm{a)}}\,\,S = \left\{ {\frac{1}{2};1} \right\}\\{\rm{b)}}\,\,x = 1\end{array}\)
C.\(\begin{array}{l}{\rm{a)}}\,\,S = \left\{ { - \frac{1}{2};1} \right\}\\{\rm{b)}}\,\,x = 2\end{array}\)
D.\(\begin{array}{l}{\rm{a)}}\,\,S = \left\{ { - \frac{1}{2}; - 1} \right\}\\{\rm{b)}}\,\,x = 3\end{array}\)
a) \(4{x^2} - 1 - \left( {1 - 2x} \right) \left( {x + 2} \right) = 0; \) b) \( \frac{{3x - {x^2}}}{{{x^2} - 9}} = 0. \)
A.\(\begin{array}{l}{\rm{a)}}\,\,S = \left\{ {\frac{1}{2}; - 1} \right\}\\{\rm{b)}}\,\,x = 0\end{array}\)
B.\(\begin{array}{l}{\rm{a)}}\,\,S = \left\{ {\frac{1}{2};1} \right\}\\{\rm{b)}}\,\,x = 1\end{array}\)
C.\(\begin{array}{l}{\rm{a)}}\,\,S = \left\{ { - \frac{1}{2};1} \right\}\\{\rm{b)}}\,\,x = 2\end{array}\)
D.\(\begin{array}{l}{\rm{a)}}\,\,S = \left\{ { - \frac{1}{2}; - 1} \right\}\\{\rm{b)}}\,\,x = 3\end{array}\)
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