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Đáp án:
Giải thích các bước giải: \(\left( {x - 3} \right)\left( {x + 1} \right) + 4\left( {x - 3} \right)\sqrt {\dfrac{{x + 1}}{{x - 3}}} + 3 = 0\) Điều kiện: \(\dfrac{{x + 1}}{{x - 3}} \ge 0 \Leftrightarrow \left[ \begin{array}{l}x \le - 1\\x > 3\end{array} \right.\) Đặt \(\left( {x - 3} \right)\sqrt {\dfrac{{x + 1}}{{x - 3}}} = t \Rightarrow {t^2} = \left( {x - 3} \right)\left( {x + 1} \right)\) Ta có phương trình \({t^2} + 4t + 3 = 0 \Leftrightarrow \left[ \begin{array}{l}t = - 1\\t = - 3\end{array} \right.\) Với \(t = - 1 \Rightarrow \left( {x - 3} \right)\sqrt {\dfrac{{x + 1}}{{x - 3}}} = - 1\) \( \Rightarrow \left\{ \begin{array}{l}x - 3 < 0\\\left( {x - 3} \right)\left( {x + 1} \right) = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x < 3\\{x^2} - 2x - 4 = 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x < 3\\{\left( {x - 1} \right)^2} = 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x < 3\\\left[ \begin{array}{l}x = 1 + \sqrt 5 \\x = 1 - \sqrt 5 \end{array} \right.\end{array} \right. \Rightarrow x = 1 - \sqrt 5 \) Với \(t = - 3 \Rightarrow \left( {x - 3} \right)\sqrt {\dfrac{{x + 1}}{{x - 3}}} = - 3\) \(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}x - 3 < 0\\\left( {x - 3} \right)\left( {x + 1} \right) = 9\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x < 3\\{x^2} - 2x - 12 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x < 3\\\left[ \begin{array}{l}x = 1 + \sqrt {13} \\x = 1 - \sqrt {13} \end{array} \right.\end{array} \right. \Rightarrow x = 1 - \sqrt {13} \end{array}\) Vậy \(x = 1 - \sqrt {13} ;x = 1 - \sqrt 5 \)