đk : x khác -2 ; 2
\(\Leftrightarrow\dfrac{x^2+3x+2-5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{12+x^2-4}{x^2-4}\)
\(\Leftrightarrow x^2-2x+12=x^2+8\Leftrightarrow-2x+4=0\Leftrightarrow x=2\left(ktm\right)\)
Vậy pt vô nghiệm
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ ĐKXĐ:x\ne2;-2\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2+4}{\left(x-2\right)\left(x+2\right)}\\ \Rightarrow x^2+2x+x+2-5x-10=12+x^2+4\\ \Leftrightarrow x^2-2x-8=16+x^2\\ \Leftrightarrow x^2-x^2-2x=16+8\\ \Leftrightarrow2x=24\\ \Leftrightarrow x=12\left(nhận\right)\)
