a) \(\dfrac{2\left(x+7\right)}{x+1}-\dfrac{x+11}{x^2-1}=4-\dfrac{x-1}{x+1}\left(ĐKXĐ:x\ne1;x\ne-1\right)\)
\(\Leftrightarrow\dfrac{2\left(x+7\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+11}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow2x^2-2x+14x-14-x-11=4x^2-4-x^2+2x-1\)
\(\Leftrightarrow2x^2+11x-25=3x^2+2x-5\)
\(\Leftrightarrow2x^2+11x-25-3x^2-2x+5=0\)
\(\Leftrightarrow-x^2+9x-20=0\)
\(\Leftrightarrow-x^2-4x-5x-20=0\)
\(\Leftrightarrow-x\left(x+4\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+\text{4}\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)(thõa mãn ĐKXĐ)
Vậy phương trình có tập nghiệm S = \(\left\{-4;-5\right\}\)
b) \(\dfrac{x+2}{x-3}+\dfrac{x+2}{x+3}=\dfrac{2\left(x+6\right)}{x^2-9}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{2\left(x+6\right)}{x^2-9}\)
\(\Rightarrow x^2+3x+2x+6+x^2-3x+2x-6=2x+12\)
\(\Leftrightarrow2x^2+4x=2x+12\)
\(\Leftrightarrow2x^2+4x-2x-12=0\)
\(\Leftrightarrow2x^2+2x-12=0\)
\(\Leftrightarrow2x^2+6x-4x-12=0\)
\(\Leftrightarrow2x\left(x+3\right)-4\left(x+3\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow x=2\)(vì x=-3 ko thõa mãn ĐKXĐ)
Vậy phương trình có tập nghiệm S = \(\left\{2\right\}\)