\(\dfrac{3x+1}{x+1}-\dfrac{2x-5}{x-3}+\dfrac{7}{x^2-2x+3}=1\left(dkxd:x\ne-1;x\ne3\right)\)
\(\Leftrightarrow\dfrac{3x+1}{x+1}-\dfrac{2x-5}{x-3}+\dfrac{7}{\left(x+1\right)\left(x-3\right)}-1=0\)
\(\Leftrightarrow\dfrac{\left(3x+1\right)\left(x-3\right)-\left(2x-5\right)\left(x+1\right)+7-\left(x+1\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow3x^2-9x+x-3-\left(2x^2+2x-5x-5\right)+7-\left(x^2-3x+x-3\right)=0\)
\(\Leftrightarrow3x^2-8x-3-2x^2+3x+5+7-x^2+2x+3=0\)
\(\Leftrightarrow-3x+12=0\)
\(\Leftrightarrow-3x=-12\)
\(\Leftrightarrow x=4\left(tmdk\right)\)
Vậy \(S=\left\{4\right\}\)
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