Question

(3/x-1)-(2/1-x)/(2/x-1)-(2/x)

Answer 1

is this (3/x-1)-[(2/1-x)/(2/x-1)]-(2/x) or [(3/x-1)-(2/1-x)]/[(2/x-1)-(2/x)] ?

Answer 2

the second one!

Answer 3

3/x(x-1)

Answer 4

by any chance, can you show me how you got that answer nubeer? You don't have to but just wondering!
i've been trying to figure these problems out for a WHILE now.

Answer 5

3/(x-1) - {(2/1-x) / 2/(x-1)} - 2/x

cancelling out the 2

3/(x-1) - {(x-1)/(1-x)} -2/x

take negative common from the second one

3/(x-1) {+(x-1)/(x-1)} -2/x

cancel x-1 which makes it 1

3/(x-1) +1 - 2/x

taking l.c.m now

{3(x) - 2(x-1) +1 } /x (x-1)

x+3 /x (x-1)

Answer 6

by the way, please try to use more brackets, this problem changes wildly depending on where you put them

Answer 7

for example, (3/x-1) is (3/x)-1 not 3/(x-1) in common notation