First look to see that x cannot equal 2 because denominators would = 0
Now change L side by getting common denominator
x^2-4 - 2x^2+4x = -1
(x^2-2x)(x-2)
the top factors to (x-2)(-x+2) then cancel out the x-2 in the top and bottom and you have
(-x+2)
(x^2-2x) = -1
or -x+2 = 2x-x^2
x^2-3x+2
(x-2)(x-1) = 0 ` solve for x = 1 or 2 (not allowed)
In the fullest details:
\(\dfrac{x + 2}{x^2 - 2x} - \dfrac2{x - 2} = -1\\ \dfrac{x + 2}{x(x - 2)} - \dfrac{2}{x - 2} = -1\\ \dfrac{x + 2}{x(x - 2)} - \dfrac{2\color{blue}x\color{black}}{\color{blue}x\color{black}(x - 2)} = -1\\ \dfrac{(x + 2) - 2x}{x(x - 2)} = -1\\ -x + 2 = -x(x - 2)\\ -x + 2 = -x^2 + 2x\\ x^2 -3x + 2 = 0\\ (x - 1)(x - 2) = 0\\ x = 1 \text{ or } x = 2\text{ (rejected)}\)