Answer 1

We have been given

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

-2 = 3x² - 6x

3x² - 6x + 2 = 0

`3x^2 - (3 + sqrt3)x-(3-sqrt3)x+3-sqrt3+sqrt3-1=0`

`x(3x-3-sqrt3)+((-3+sqrt3)/3)(3x-3-sqrt3)=0`

`((3x-3+sqrt3)/3)(3x-3-sqrt3)=0`

`(sqrt3x-sqrt3+1)(sqrt3x-sqrt3-1)=0`

Therefore,

`sqrt3x-sqrt3+1=0`

`sqrt3x=sqrt3-1`

`x(sqrt3-1)/sqrt3`

or

`sqrt3x-sqrt3-1=0`

`sqrt3x=sqrt3+1`

`x(sqrt3+1)/sqrt3`

Hence, `x(sqrt3-1)/sqrt3` or `x(sqrt3+1)/sqrt3`