Lời giải:
ĐKXĐ: $x\neq -1;-2;0;-11$
PT \(\Leftrightarrow \frac{3}{x^2+3x+2}-\frac{6}{x^2+11x}+\frac{2}{x^2-x+3}-\frac{8}{x^2+11x}=0\)
\(\Leftrightarrow \frac{-3x^2+15x-12}{(x^2+3x+2)(x^2+11x)}+\frac{-6x^2+30x-24}{(x^2-x+3)(x^2+11x)}=0\)
\(\Leftrightarrow \frac{-3x^2+15x-12}{x^2+11x}\left(\frac{1}{x^2+3x+2}+\frac{2}{x^2-x+3}\right)=0\)
\(\Leftrightarrow \frac{-3x^2+15x-12}{x^2+11x}.\frac{3x^2+5x+7}{(x^2+3x+2)(x^2-x+3)}=0\)
\(\Rightarrow (-3x^2+15x-12)(3x^2+5x+7)=0\)
\(\Rightarrow x=1\) hoặc $x=4$ (thỏa mãn)
Vậy.....
