a) đkxđ: \(x\ne\pm2\)
b) \(A=\left(\dfrac{x+2}{2x-4}-\dfrac{2-x}{2x+4}+\dfrac{8}{x^2-4}\right):\dfrac{x-1}{x-2}\\ =\left(\dfrac{\left(x+2\right)^2}{2\left(x+2\right)\left(x-2\right)}+\dfrac{\left(x-2\right)^2}{2\left(x+2\right)\left(x-2\right)}+\dfrac{16}{2\left(x+2\right)\left(x-2\right)}\right):\dfrac{x-1}{x-2}\\ =\dfrac{\left(x+2\right)^2+\left(x-2\right)^2+16}{2\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x-2}{x-1}\\ =\dfrac{x^2+4x+4+x^2-4x+4+16}{2\left(x+2\right)}\cdot\dfrac{1}{x-1}\\ =\dfrac{2x^2+24}{2\left(x+2\right)\left(x-1\right)}\\ =\dfrac{x^2+12}{\left(x+2\right)\left(x-1\right)}\)
c) \(A=1\Leftrightarrow\dfrac{x^2+12}{\left(x-2\right)\left(x-1\right)}=1\\ \Leftrightarrow x^2+12=\left(x-2\right)\left(x-1\right)\\ \Leftrightarrow x^2+12=x^2-3x+2\\ \Leftrightarrow12=-3x+2\\ \Leftrightarrow3x=-10\\ \Leftrightarrow x=-\dfrac{10}{3}\)
