#(1) / (x -2 ) = (1) + (2) / (x^2-2x)#
There are some restrictions on what values of #x# we can use. Some people like to state those restrictions at the beginning and later check to see that we have not violated those restrictions. My habit is to just check at the end.
Notice that #x^2 - 2x = x(x-2)# so we can clear the denominators be multiplying through by that:
#(1/(x -2 )) * (x(x-2))/1= (1 + 2/(x^2-2x))(x^2 - 2x )/1#
#x = x^2-2x + 2#
#x^2 - 3x +2 = 0#
#(x-2)(x-1)=0#
#x=2" "# or #" "x=1#
Check for extraneous solutions:
#x=2# makes one (at least one, in fact two) of the denominators equal to #0#, so that is NOT a solution to the original equation.
#x=1# gets us:
#-1 = 1-2#
which is true. So #x=1# is a solution. (The only solution.)
