Sửa đề: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10-12-x^2+4=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
\(\Leftrightarrow\) \(\dfrac{x+1}{x+2}-\dfrac{5}{x+2}-\dfrac{12}{\left(x+2\right)\left(x-2\right)}-1=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{12}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=0\) 0
\(\Leftrightarrow x^2+x-2x-2-5x+10-12-x^2+4=0\)\(\Leftrightarrow\)\(-6x=0\Leftrightarrow x=0\)