Đáp án:
c. \(\left[ \begin{array}{l}
a = 0\\
a = \dfrac{1}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Thay:a = - 3\\
Pt \to \dfrac{{x + 3}}{{x - 3}} - \dfrac{{x - 3}}{{x + 3}} + \dfrac{{24}}{{{x^2} - 9}} = 0\left( {DK:x \ne \pm 3} \right)\\
\to \dfrac{{{x^2} + 6x + 9 - {x^2} + 6x - 9 + 24}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = 0\\
\to 12x + 24 = 0\\
\to x = - 2\left( {TM} \right)\\
b.Thay:a = 1\\
Pt \to \dfrac{{x - 1}}{{x + 1}} - \dfrac{{x + 1}}{{x - 1}} + \dfrac{4}{{{x^2} - 1}} = 0\left( {DK:x \ne \pm 1} \right)\\
\to \dfrac{{{x^2} - 2x + 1 - {x^2} - 2x - 1 + 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
\to - 4x + 4 = 0\\
\to x = 1\left( l \right)\\
\to x \notin \emptyset \\
c.Thay:x = \dfrac{1}{2}\\
Pt \to \dfrac{{\dfrac{1}{2} - a}}{{\dfrac{1}{2} + a}} - \dfrac{{\dfrac{1}{2} + a}}{{\dfrac{1}{2} - a}} + \dfrac{{3{a^2} + a}}{{\dfrac{1}{4} - {a^2}}} = 0\left( {DK:a \ne \pm \dfrac{1}{2}} \right)\\
\to \dfrac{1}{4} - a + {a^2} - \dfrac{1}{4} - a - {a^2} + 3{a^2} + a = 0\\
\to 3{a^2} - a = 0\\
\to a\left( {3a - 1} \right) = 0\\
\to \left[ \begin{array}{l}
a = 0\\
a = \dfrac{1}{3}
\end{array} \right.\left( {TM} \right)
\end{array}\)
