a) \(ĐKXĐ:\)\(x\ne1;\)\(x\ne2;\)\(x\ne3.\)
\(\frac{6}{x^2-3x+2}+\frac{4}{x^2-4x+3}=\frac{2}{x^2-5x+6}\)
\(\Leftrightarrow\)\(\frac{6}{\left(x-1\right)\left(x-2\right)}+\frac{4}{\left(x-1\right)\left(x-3\right)}=\frac{2}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow\)\(\frac{6\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{4\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\Rightarrow\)\(6\left(x-3\right)+4\left(x-2\right)=2\left(x-1\right)\)
\(\Leftrightarrow\)\(6x-18+4x-8=2x-2\)
\(\Leftrightarrow\)\(8x=24\)
\(\Leftrightarrow\)\(x=3\) (ko thỏa mãn ĐKXĐ)
Vậy pt vô nghiệm