\(1)\dfrac{x+4}{x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\\ \Leftrightarrow\dfrac{x+4}{x^2-5x+2}+\left(\dfrac{x+1}{2x^2-7x+3}-\dfrac{2x+5}{2x^2-7x+3}\right)=0\\ \Leftrightarrow\dfrac{x+4}{x^2-5x+2}-\dfrac{x+4}{2x^2-7x+3}=0\\ \Leftrightarrow\left(x+4\right)\left(\dfrac{1}{x^2-5x+2}-\dfrac{1}{2x^2-7x+3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+4=0\\\dfrac{1}{x^2-5x+2}-\dfrac{1}{2x^2-7x+3}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\2x^2-7x+3=x^2-5x+2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x^2-2x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
\(2)\dfrac{2}{x^2+2x+1}-\dfrac{5}{x^2-2x+1}=\dfrac{3}{1-x^2}\left(x\ne\pm1\right)\\ \Leftrightarrow\dfrac{2}{\left(x+1\right)^2}-\dfrac{5}{\left(x-1\right)^2}=\dfrac{-3}{x^2-1}\\ \Leftrightarrow\dfrac{2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}-\dfrac{5\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=\dfrac{-3\left(x+1\right)\left(x-1\right)}{\left(x+1\right)^2\left(x-1\right)^2}\\ \Leftrightarrow2\left(x-1\right)^2-5\left(x+1\right)^2=-3\left(x+1\right)\left(x-1\right)\\ \Leftrightarrow2\left(x^2-2x+1\right)-5\left(x^2+2x+1\right)=-3\left(x^2-1\right)\\ \Leftrightarrow2x^2-4x+2-5x^2-10x-5=-3x^2+3\\ \Leftrightarrow-3x^2-14x-3=-3x^2+3\\ \Leftrightarrow-3x^2+3x^2=14x+3+3\\ \Leftrightarrow14x+6=0\\ \Leftrightarrow14x=-6\\ \Leftrightarrow x=-\dfrac{6}{14}\\ \Leftrightarrow x=\dfrac{-3}{7}\left(tm\right)\)
