\(\dfrac{3}{\left(x-1\right)\left(x-3\right)}+\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{\left(x-2\right)\left(x-3\right)}\)( ĐKXĐ: \(x\ne1\); \(x\ne3\); \(x\ne2\))
\(\Leftrightarrow\dfrac{3+2}{\left(x-1\right)\left(x-3\right)}=\dfrac{1}{\left(x-3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{5\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\Rightarrow5x-10=x-1\)
\(\Leftrightarrow5x-x=10-1\)
\(\Leftrightarrow4x=9\)
\(\Leftrightarrow x=\dfrac{9}{4}\)( thõa mãn ĐKXĐ)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{9}{4}\right\}\)
