`a, ĐKXĐ: {(2x-4 ne 0),(2x+4 ne 0),(x^2 -4 ne 0),( x-1 ne 0),(x-2 ne 0):}`
`<=> {(x ne 2),(x ne -2),(x ne +- 2),( x ne 1),(x ne 2):}`
`<=> {(x ne +- 2),(x ne 1):}`
\(b,A=\left(\dfrac{x+2}{2x-4}-\dfrac{2-x}{2x+4}+\dfrac{8}{x^2-4}\right):\dfrac{x-1}{x-2}\\ =\left(\dfrac{x+2}{2\left(x-2\right)}-\dfrac{2-x}{2\left(x+2\right)}+\dfrac{8}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x-2}{x-1}\\ =\left(\dfrac{\left(x+2\right)^2}{2\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)^2}{2\left(x+2\right)\left(x-2\right)}+\dfrac{2.8}{2\left(x-2\right)\left(x+2\right)}\right).\dfrac{x-2}{x-1}\\ =\left(\dfrac{x^2+4x+4}{2\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4x+4}{2\left(x+2\right)\left(x-2\right)}+\dfrac{16}{2\left(x-2\right)\left(x+2\right)}\right).\dfrac{x-2}{x-1}\\ =\dfrac{x^2+4x+4+x^2-4x+4+16}{2\left(x-2\right)\left(x+2\right)}.\dfrac{x-2}{x-1}\)
\(=\dfrac{2x^2+24}{2\left(x+2\right)\left(x-1\right)}\\ =\dfrac{2\left(x^2+12\right)}{2\left(x+2\right)\left(x-1\right)}\\ =\dfrac{x^2+12}{\left(x+2\right)\left(x-1\right)}\)
\(c,A=1\\ \Leftrightarrow\dfrac{x^2+12}{\left(x+2\right)\left(x-1\right)}=1\\ \Leftrightarrow\dfrac{x^2+12}{x^2+2x-x-2}=1\\ \Leftrightarrow\dfrac{x^2+12}{x^2+x-2}=1\\ \Rightarrow x^2+12=x^2+x-2\\ \Leftrightarrow x^2+x-2-x^2-12=0\\ \Leftrightarrow x-14=0\\ \Leftrightarrow x=14\left(tm\right)\)
