Đáp án:
$\begin{array}{l}
a)Đkxđ:\left\{ \begin{array}{l}
x + 3 \ne 0\\
x - 3 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne - 3\\
x \ne 3
\end{array} \right.\\
\frac{{x + 3}}{{x + 3}} + \frac{{x - 3}}{{x - 3}} = 2\\
\Rightarrow 1 + 1 = 2\left( {luon\,dung} \right)
\end{array}$
Vậy pt có nghiệm đúng với mọi x khác 3 và khác -3
b)
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x - 1 \ne 0\\
x - 3 \ne 0\\
{x^2} - 4x + 3 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne 3
\end{array} \right.\\
\frac{{x + 5}}{{x - 1}} = \frac{{x + 1}}{{x - 3}} - \frac{8}{{{x^2} - 4x + 3}}\\
\Rightarrow \frac{{x + 5}}{{x - 1}} = \frac{{\left( {x + 1} \right)\left( {x - 1} \right) - 8}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\\
\Rightarrow \frac{{x + 5}}{{x - 1}} = \frac{{{x^2} - 1 - 8}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\\
\Rightarrow \frac{{x + 5}}{{x - 1}} = \frac{{{x^2} - 9}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\\
\Rightarrow x + 5 = \frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{x - 3}}\\
\Rightarrow x + 5 = x + 3\\
\Rightarrow 5 = 3\left( {ktm} \right)
\end{array}$
Vậy pt vô nghiệm.
