a)ĐKXĐ: x≠-1; x≠3
Ta có: \(\frac{3x+1}{x+1}-\frac{2x-5}{x-3}+\frac{7}{x^2-2x-3}=1\)
\(\Leftrightarrow\frac{3x+1}{x+1}-\frac{2x-5}{x-3}+\frac{7}{\left(x+1\right)\left(x-3\right)}-1=0\)
\(\Leftrightarrow\frac{\left(3x+1\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}-\frac{\left(2x-5\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\frac{7}{\left(x+1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow3x^2-8x-3-\left(2x^2-3x-5\right)+7-\left(x^2-2x-3\right)=0\)
\(\Leftrightarrow3x^2-8x-3-2x^2+3x+5+7-x^2+2x+3=0\)
\(\Leftrightarrow-3x+12=0\)
\(\Leftrightarrow-3x=-12\)
hay x=4
Vậy: x=4
b) ĐKXĐ: x≠-2; x≠2
Ta có: \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x^2-4x+4-\left(3x+6\right)-\left(2x-22\right)=0\)
\(\Leftrightarrow x^2-4x+4-3x-6-2x+22=0\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy: x∈{4;5}