fractional equation in algebra 1?

2/c^2-2c + 1/2-c = 1
....2..............1
------------ + -----------=1
c(c-2)..........2-c

....2...............1
---------- +--------------=1
-c(2-c)..........2-c

....1............1
---------- - -----------=1 lcd:c( 2-c)
2-c........c(2-c)

c-1=c(2-c)
c-2=2c-c^2
c-2c+c^2-2=0
c^2-c-2=0
(c+1)(c-2)=0
c+1=0 c-2=0
c=-1 c=2 answer

well first of all you need to clarify your question more clearly so it is easier to understand. im guessing ur trying to say what is "c" well first off all you need to rearrange your equation to get "c" by its self ill show you how to do it step wise
(2^2/C^2) - 2C + 1/2 - C = 1
you need to add all -C + -3 which gives you
(2^2/C^2) - 3C + 1/2 -1 = 0
then you must add 1/2 with -1 which gives you
(2^2/C^2) - 3C - 1/2 = 0
then you add the 2^2
(4/C^2) - 3C -1 = 0
and then you can just rearrange that into the quadratic formula x= -b(+-)Square root b^2 - 4ac/2a
where a=4 b= -3 and c=-1 and just rearrange them into your equation and that should give you your answer whether it be in surd form im not sure

2/(c^2 - 2c) + 1/(2 - c) = 1
2/c(c - 2) + 1/-(-2 + c) = 1
2/c(c - 2) - 1/(c - 2) = 1
c[c - 2][2/c(c - 2) - 1/(c - 2)] = c[c - 2][1]
2 - c = c*c - c*2
2 - c = c^2 - 2c
c^2 - 2c + c - 2 = 0
c^2 - c - 2 = 0
c^2 + c - 2c - 2 = 0
(c^2 + c) - (2c + 2) = 0
c(c + 1) - 2(c + 1) = 0
(c + 1)(c - 2) = 0

c + 1 = 0
c = -1

c - 2 = 0
c = 2

∴ c = -1 , 2

You must present the question unambiguously.

What's the meaning of your question? is it 2 divided by c to the power of (2-2c), or 2 divided by (c^2-2c), or 2 divided by c^2, then -2c is outside?