Cho phương trình \(\dfrac{{x - 2}}{{x + 3}} + 1 = \dfrac{{3x - 1}}{{x - 3}}\). Khẳng định nào sau đây là sai?

Điều kiện xác định: \(\left\{ \begin{array}{l}x + 3 \ne 0\\x - 3 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne - 3\\x \ne 3\end{array} \right.\)

\(\begin{array}{l}\dfrac{{x - 2}}{{x + 3}} + 1 = \dfrac{{3x - 1}}{{x - 3}}\\ \Leftrightarrow \dfrac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} + \dfrac{{\left( {x + 3} \right)\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{\left( {3x - 1} \right)\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\ \Leftrightarrow \left( {x - 2} \right)\left( {x - 3} \right) + \left( {x + 3} \right)\left( {x - 3} \right) = \left( {3x - 1} \right)\left( {x + 3} \right)\\ \Leftrightarrow {x^2} - 5x + 6 + {x^2} - 9 - \left( {3{x^2} + 8x - 3} \right) = 0\\ \Leftrightarrow {x^2} - 5x + 6 + {x^2} - 9 - 3{x^2} - 8x + 3 = 0\\ \Leftrightarrow {x^2} + 13x = 0\\ \Leftrightarrow x\left( {x + 13} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x + 13 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\left( {tm} \right)\\x = - 13\left( {tm} \right)\end{array} \right.\end{array}\)

⇒ Tổng hai nghiệm bằng -13