\(\dfrac{x-1}{x-2}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow x^2+x-2+5x-10=12+x^2-4\)
\(\Leftrightarrow6x-12=8\)
=>6x=20
hay x=10/3(nhận)
ĐKXĐ:\(x\ne\pm2\)
\(\dfrac{1-x}{2-x}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{12}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{x^2+x-2+5x-10-12-x^2+4}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow6x-20=0\\ \Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)
