\(\Rightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+10x+8x+80}=\frac{9}{52}\)
\(\Rightarrow\frac{2}{x\left(x+1\right)+3\left(x+1\right)}+\frac{5}{x\left(x+3\right)+8\left(x+3\right)}+\frac{2}{x\left(x+10\right)+8\left(x+10\right)}=\frac{9}{52}\)
\(\Rightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)
\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\Rightarrow\frac{x+10-x-1}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\Rightarrow\frac{9}{x^2+11x+10}=\frac{9}{52}\)
\(\Rightarrow x^2+11x+10=52\Rightarrow x^2+2\cdot\frac{11}{2}x+\frac{121}{4}-\frac{81}{4}=52\)
\(\Rightarrow\left(x+\frac{11}{2}\right)^2=\frac{289}{4}\Rightarrow x+\frac{11}{2}=\frac{17}{2}\Rightarrow x=\frac{17}{2}-\frac{11}{2}=\frac{6}{2}=3\Rightarrow x=3\)
\(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{52}\)(ĐKXĐ: x khác -1;-3;-8;-10)
\(\Leftrightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+8x+10x+80}=\frac{9}{52}\)
\(\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\frac{2\left(x+8\right)\left(x+10\right)+5\left(x+1\right)\left(x+10\right)+2\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\frac{9x^2+99x+216}{x^4+22x^3+155x^2+374x+240}=\frac{9}{52}\)
\(\Rightarrow468x^2+5148x+11232=9x^4+198x^3+1395x^2+3366x+2160\)
\(\Leftrightarrow9x^4+198x^3+927x^2-1782x-9072=0\)
\(\Leftrightarrow x^4+22x^3+103x^2-198x-1008=0\)
\(\Leftrightarrow x^4-3x^3+25x^3-75x^2+178x^2-534x+336x-1008=0\)
\(\Leftrightarrow x^3\left(x-3\right)+25x^2\left(x-3\right)+178x\left(x-3\right)+336\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3+25x^2+178x+336\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2+22x^2+66x+112x+336\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[x^2\left(x+3\right)+22x\left(x+3\right)+112\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+22x+112\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+8x+14x+112\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left[x\left(x+8\right)+14\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left(x+8\right)\left(x+14\right)=0\)
\(\Leftrightarrow\frac{\orbr{\begin{cases}x+3=0\\x-3=0\end{cases}}}{\orbr{\begin{cases}x+8=0\\x+14=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-3\left(\times\right)\\x=3\end{cases}}}{\orbr{\begin{cases}x=-8\left(\times\right)\\x=-14\end{cases}}}\)(Vì x=-3 và x=-8 không t/m ĐKXĐ)
Vậy tập nghiệm của pt là \(S=\left\{3;-14\right\}.\)
