\(\dfrac{2}{x-1}-\dfrac{x-3}{x^2-2x+1}=0\left(x\ne1\right)\\ < =>\dfrac{2}{x-1}-\dfrac{x-3}{\left(x-1\right)^2}=0\)
suy ra
\(2\left(x-1\right)-x+3=0\\ < =>2x-2-x+3=0\)
\(< =>x+1=0\\ < =>x=-1\left(tm\right)\)
\(\dfrac{2}{x-1}-\dfrac{x-3}{x^2-2x+1}=0\text{ĐKXĐ:}x\ne1\)
\(\Leftrightarrow\dfrac{2}{x-1}-\dfrac{x-3}{\left(x-1\right)^2}=0\)
\(\Leftrightarrow\dfrac{2\left(x-1\right)}{\left(x-1\right)^2}-\dfrac{x-3}{\left(x-1\right)^2}=0\)
\(\Rightarrow2\left(x-1\right)-x+3=0\)
\(\Leftrightarrow2x-2-x+3=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\left(TM\text{ĐKXĐ}\right)\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{-1\right\}\)
