ĐKXĐ của phương trình là :
\(\left\{{}\begin{matrix}x^2+4x+3\ne0\\x^2+8x+15\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(x+3\right)\ne0\\\left(x+3\right)\left(x+5\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-3\\x\ne-5\end{matrix}\right.\)
\(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{x^2+3x+x+3}+\dfrac{1}{x^2+3x+5x+15}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{x+5}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}+\dfrac{x+1}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{2x+6}{\left(x+1\right)\left(x+3\right)\left(x+6\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{2\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+5\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=12\)
\(\Leftrightarrow x^2+6x+5=12\)
\(\Leftrightarrow x^2+6x+5-12=0\)
\(\Leftrightarrow x^2+6x-7=0\)
\(\Leftrightarrow x^2-7x+x-7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\) ( t/m )
Vậy tập nghiệm của pt là \(S=\left\{-1;7\right\}\)
