a/ ĐK: x ≠ \(\pm\)2; x≠1
\(A=\left(\dfrac{x+2}{2x-4}-\dfrac{2-x}{2x+4}+\dfrac{8}{x^2-4}\right):\dfrac{x-1}{x-2}\)
= (\(\dfrac{x+2}{2\left(x-2\right)}-\dfrac{2-x}{2\left(x+2\right)}+\dfrac{8}{\left(x-2\right)\left(x+2\right)}\)) \(\cdot\dfrac{x-2}{x-1}\)
\(=\left(\dfrac{\left(x+2\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{2\cdot8}{2\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x-2}{x-1}\)
\(=\left(\dfrac{x^2+4x+4+x^2-4x+4+16}{2\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x-2}{x-1}\)
\(=\dfrac{2x^2+24}{2\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x-2}{x-1}\)
\(=\dfrac{2\left(x^2+12\right)}{2\left(x+2\right)}\cdot\dfrac{1}{x-1}=\dfrac{x^2+12}{\left(x+2\right)\left(x-1\right)}\)
b/ \(A=1\Leftrightarrow\dfrac{x^2+12}{\left(x+2\right)\left(x-1\right)}=1\)
\(\Leftrightarrow x^2+12=\left(x+2\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+12=x^2+x-2\)
\(\Leftrightarrow-x=-14\Leftrightarrow x=14\)(t/m)
Vậy..............................
