ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x+2\right)\left(x-2\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x+2\right)\left(x-2\right)\\ \Leftrightarrow x^2+x+2x+2-5x+10=12+x^2-4\\ \Leftrightarrow-2x=-4\\ \Leftrightarrow x=2\left(ktm\right)\)
Vậy \(S\in\left\{\varnothing\right\}\)
ĐKXĐ: \(\begin{cases}x-2\ne 0\\x+2\ne 0\end{cases}\leftrightarrow x\ne 2\\x\ne -2\end{cases}\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\leftrightarrow \dfrac{(x+1)(x+2)}{(x-2)(x+2)}-\dfrac{5(x-2)}{(x+2)(x-2)}=\dfrac{12}{(x-2)(x+2)}+\dfrac{(x-2)(x+2)}{(x-2)(x+2)}\)
\(\to x^2+3x+2-5x+10=12+x^2-4\)
\(\leftrightarrow x^2-2x-x^2=12-12-4\)
\(\leftrightarrow -2x=-4\)
\(\leftrightarrow x=2(\rm KTM)\)
Vậy pt đã cho vô nghiệm \(S=\varnothing\)
