Giải các phương trình:
a) \(\frac{{x + 3}}{{x - 3}} - \frac{{x - 3}}{{x + 3}} = \frac{{12}}{{{x^2} - 9}}\) b) \(\left| {x - 3} \right| = 4x + 9\)
A.\(\begin{array}{l}a)\,\,x = 1\\b)\,\,x = - \frac{6}{5}\end{array}\)
B.\(\begin{array}{l}a)\,\,x = - 1\\b)\,\,\left[ \begin{array}{l}x = - 4\\x = - \frac{6}{5}\end{array} \right.\end{array}\)
C.\(\begin{array}{l}a)\,\,x = 1\\b)\,\,\left[ \begin{array}{l}x = 4\\x = \frac{6}{5}\end{array} \right.\end{array}\)
D.\(\begin{array}{l}a)\,\,x = - 1\\b)\,\,x = \frac{6}{5}\end{array}\)
a) \(\frac{{x + 3}}{{x - 3}} - \frac{{x - 3}}{{x + 3}} = \frac{{12}}{{{x^2} - 9}}\) b) \(\left| {x - 3} \right| = 4x + 9\)
A.\(\begin{array}{l}a)\,\,x = 1\\b)\,\,x = - \frac{6}{5}\end{array}\)
B.\(\begin{array}{l}a)\,\,x = - 1\\b)\,\,\left[ \begin{array}{l}x = - 4\\x = - \frac{6}{5}\end{array} \right.\end{array}\)
C.\(\begin{array}{l}a)\,\,x = 1\\b)\,\,\left[ \begin{array}{l}x = 4\\x = \frac{6}{5}\end{array} \right.\end{array}\)
D.\(\begin{array}{l}a)\,\,x = - 1\\b)\,\,x = \frac{6}{5}\end{array}\)
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