\(\dfrac{x-1}{x+2}-\dfrac{x}{x+2}=\dfrac{2-5x}{x^2-4}\\ ĐKXĐ:x\ne-2;x\ne2\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2-5x}{\left(x+2\right)\left(x-2\right)}\\ \Rightarrow x^2-2x-x+2-x^2+2x=2-5x\\ \Leftrightarrow-x+2=2-5x\\ \Leftrightarrow-x+5x=2-2\\ \Leftrightarrow4x=0\\ \Leftrightarrow x=0\left(nhận\right)\)